##### Class 10^{th} Mathematics Gujarat Board Solution

**Exercise 2.1**- Identify the type of the following polynomials: (on base of power) (1) p(x) =…
- Obtain the degree of the following polynomials: (1) p(x) = 3x —x^4 + x^2 + 2x^3…
- Find the coefficients of the underlined terms: (1) p(x) = 10x^3 + 7 x^2 — 3x + 5…
- Obtain the value of the following polynomials at the given values of x: (1) p(x)…
- (x + 1) is a factor of p(x) = 3x^3 + 2x^2 + 7 x + 8 Examine the validity of the…
- (x + 2) is a factor of p(x) = x^3 + x^2 + x + 2 Examine the validity of the…
- (x — 1) is a factor of p(x) = x^4 — 2x^3 + 3x —2 Examine the validity of the…
- (x — 3) is a factor of p(x) = x^2 — 2x — 3 Examine the validity of the…
- p(x) = x^3 —x^2 —x + 1 Factorize the following polynomials:
- p(x) = 5x^2 + 11x + 6 Factorize the following polynomials:
- p(x) = x^3 — 3x^2 + 9x — 27 Factorize the following polynomials:
- p(x) = x^3 + 2x^2 + 3x + 2 Factorize the following polynomials:
- Prove that x — 2 is a factor of p(x) = x^3 — 2x^2

**Exercise 2.2**- p(x) = x^2 — x Find the number of zeros of the following polynomials:…
- p(x) = x —x^2 —1 Find the number of zeros of the following polynomials:…
- P(x) = 3x —2 Find the number of zeros of the following polynomials:…
- p(x) = x^3 — x Find the number of zeros of the following polynomials:…
- Find the number of zeros and real zeros of p(x) = x^3 + 1. Show them by a graph.…
- Draw the graph of p(x) = x^2 + 1 and find the real zeros of this polynomial.…
- From the figure 2.10 find the number of zeros of y = p(x). (i) (ii) (iii) (iv) (v) (vi)…
- Find the number of zeros and zeros of p(x) = x^2 — 4. Represent them…

**Exercise 2.3**- Prove that 4 and 1 are the zeros of the quadratic polynomial p(x) = x^2 — 5x +…
- p(x) = x^2 + 4x — 21 Find the zeros of the following quadratic polynomials:…
- p(x) = 6x^2 — 11x + 5 Find the zeros of the following quadratic polynomials:…
- p(x) = 4x^2 + 9x + 5 Find the zeros of the following quadratic polynomials:…
- p(x) = 3x^2 + 5x — 8 Find the zeros of the following quadratic polynomials:…
- p(x) = x^2 — 81 Find the zeros of the following quadratic polynomials:…
- p(x) = x^2 — x — 6 Find the zeros of the following quadratic polynomials:…
- Find the zeros, the sum of the zeros and the product of the zeros of the…
- The sum of zeros = 2; the product of zeros = —3 Obtain a quadratic polynomial…
- The sum of zeros = —3; the product of zeros = —4 Obtain a quadratic polynomial…
- The sum of zeros = 1/3; the product of zeros = 1/2 Obtain a quadratic…
- Obtain the quadratic or the cubic polynomial as the case may be in the standard…

**Exercise 2.4**- p(x) = 2x^3 — 13x^2 + 23x — 12, s(x) = 2x — 3 Divide the following polynomial…
- p(x) = 2/3 x^2 + 5x + 6, s(x) = x + 6 Divide the following polynomial p(x) by…
- p(x) = 40x^2 + 11x — 63, s(x) = 8x — 9 Divide the following polynomial p(x) by…
- p(x) = 2x^3 + 9x^2 + 13x + 6, s(x) = 2x^2 + 5x + 3 Divide the following…
- p(x) = x^4 + 4x^3 + 5x^2 — 7x — 3, s(x) = x^2 — 1 Divide the following…
- Find the remainder polynomial when the cubic polynomial x^3 — 3x^2 + 4x + 5 is…
- 3 is a zero of p(x) = 3x^3 — x^2 — ax — 45. Find 'a'.
- The product of two polynomials is 6x^3 + 29x^2 + 44x + 21 and one of the…
- If polynomial p(x) is divided by x^2 + 3x + 5, the quotient polynomial and the…
- Divide p(x) = x^3 — 4x^2 + 5x — 2 by x — 2 Find r(x).
- There are x^4 + 57x + 15 pens to be distributed in a class of x^2 + 4x + 2…
- A trader bought 2x^2 — x + 2 TV sets for Rs. 8x^4 + 7x — 6. Find the price of…
- - root 2 and root 2 are two of the zeros of p(x) = 2x^4 + 7x^3 — 8x^2 — 14x + 8.…

**Exercise 2**- State whether the following statements are true or false: (1) 7/5 is a zero of…
- Find the zeros and number of zeros of p(x) = x^2 + 9x + 18. Show them on a graph.…
- Find the zeros, the sum and the product of zeros of p(x) = 4x^2 + 12x + 5.…
- —4 and 9 are the sum and product of the zeros respectively of a quadratic…
- Find q(x) and r(x), for the quadratic polynomial p(x) = 11x — 21 + 2x^2 when…
- Divide 2x^3 + 3x^2 — 11x — 6 by x^2 + x — 6
- 4 is a zero of the cubic polynomial p(x) = x^3 — 3x^2 — 6x + 8. Find the…
- The product of two polynomials is 3x^4 + 5x^3 — 21x^2 — 53x — 30. If one of them…
- 2 + √3 and 2 — √3 are the zeros of p(x) = x^4 — 6x^3 — 26x^2 + 138x — 35. Find…
- The linear polynomial p(x) = 7x — 3 has the zero _____A. 7/3 B. 3/7 C. - 7/3…
- The cubic polynomial p(x) = x^3 — x has _____ zeros.A. 0 B. 1 C. 2 D. 3…
- The graph of p(x) = 3x — 2 — x^2 intersects the X-axis in _____ points.A. 0 B.…
- The sum of the zeros of 3x^2 + 5x — 2 is _____A. 3/5 B. - 3/5 C. 5/3 D. - 5/3…
- The graph of p(x) = 3x + 5 represents _____A. a straight line B. parabola open…
- A quadratic polynomial has no zero. Its graph _____A. touches X-axis at any…
- For the graph in figure 2.11 y = p(x) has _____ zeros. A. 1 B. 2 C. 3 D. 4…
- The product of the zeros of x^2 — 4x + 3 is _____A. 1 B. 3 C. 4 D. —4…
- a = 3, b = 5, c = 7, d = 11 in the standard notation gives the cubic…

**Exercise 2.1**

- Identify the type of the following polynomials: (on base of power) (1) p(x) =…
- Obtain the degree of the following polynomials: (1) p(x) = 3x —x^4 + x^2 + 2x^3…
- Find the coefficients of the underlined terms: (1) p(x) = 10x^3 + 7 x^2 — 3x + 5…
- Obtain the value of the following polynomials at the given values of x: (1) p(x)…
- (x + 1) is a factor of p(x) = 3x^3 + 2x^2 + 7 x + 8 Examine the validity of the…
- (x + 2) is a factor of p(x) = x^3 + x^2 + x + 2 Examine the validity of the…
- (x — 1) is a factor of p(x) = x^4 — 2x^3 + 3x —2 Examine the validity of the…
- (x — 3) is a factor of p(x) = x^2 — 2x — 3 Examine the validity of the…
- p(x) = x^3 —x^2 —x + 1 Factorize the following polynomials:
- p(x) = 5x^2 + 11x + 6 Factorize the following polynomials:
- p(x) = x^3 — 3x^2 + 9x — 27 Factorize the following polynomials:
- p(x) = x^3 + 2x^2 + 3x + 2 Factorize the following polynomials:
- Prove that x — 2 is a factor of p(x) = x^3 — 2x^2

**Exercise 2.2**

- p(x) = x^2 — x Find the number of zeros of the following polynomials:…
- p(x) = x —x^2 —1 Find the number of zeros of the following polynomials:…
- P(x) = 3x —2 Find the number of zeros of the following polynomials:…
- p(x) = x^3 — x Find the number of zeros of the following polynomials:…
- Find the number of zeros and real zeros of p(x) = x^3 + 1. Show them by a graph.…
- Draw the graph of p(x) = x^2 + 1 and find the real zeros of this polynomial.…
- From the figure 2.10 find the number of zeros of y = p(x). (i) (ii) (iii) (iv) (v) (vi)…
- Find the number of zeros and zeros of p(x) = x^2 — 4. Represent them…

**Exercise 2.3**

- Prove that 4 and 1 are the zeros of the quadratic polynomial p(x) = x^2 — 5x +…
- p(x) = x^2 + 4x — 21 Find the zeros of the following quadratic polynomials:…
- p(x) = 6x^2 — 11x + 5 Find the zeros of the following quadratic polynomials:…
- p(x) = 4x^2 + 9x + 5 Find the zeros of the following quadratic polynomials:…
- p(x) = 3x^2 + 5x — 8 Find the zeros of the following quadratic polynomials:…
- p(x) = x^2 — 81 Find the zeros of the following quadratic polynomials:…
- p(x) = x^2 — x — 6 Find the zeros of the following quadratic polynomials:…
- Find the zeros, the sum of the zeros and the product of the zeros of the…
- The sum of zeros = 2; the product of zeros = —3 Obtain a quadratic polynomial…
- The sum of zeros = —3; the product of zeros = —4 Obtain a quadratic polynomial…
- The sum of zeros = 1/3; the product of zeros = 1/2 Obtain a quadratic…
- Obtain the quadratic or the cubic polynomial as the case may be in the standard…

**Exercise 2.4**

- p(x) = 2x^3 — 13x^2 + 23x — 12, s(x) = 2x — 3 Divide the following polynomial…
- p(x) = 2/3 x^2 + 5x + 6, s(x) = x + 6 Divide the following polynomial p(x) by…
- p(x) = 40x^2 + 11x — 63, s(x) = 8x — 9 Divide the following polynomial p(x) by…
- p(x) = 2x^3 + 9x^2 + 13x + 6, s(x) = 2x^2 + 5x + 3 Divide the following…
- p(x) = x^4 + 4x^3 + 5x^2 — 7x — 3, s(x) = x^2 — 1 Divide the following…
- Find the remainder polynomial when the cubic polynomial x^3 — 3x^2 + 4x + 5 is…
- 3 is a zero of p(x) = 3x^3 — x^2 — ax — 45. Find 'a'.
- The product of two polynomials is 6x^3 + 29x^2 + 44x + 21 and one of the…
- If polynomial p(x) is divided by x^2 + 3x + 5, the quotient polynomial and the…
- Divide p(x) = x^3 — 4x^2 + 5x — 2 by x — 2 Find r(x).
- There are x^4 + 57x + 15 pens to be distributed in a class of x^2 + 4x + 2…
- A trader bought 2x^2 — x + 2 TV sets for Rs. 8x^4 + 7x — 6. Find the price of…
- - root 2 and root 2 are two of the zeros of p(x) = 2x^4 + 7x^3 — 8x^2 — 14x + 8.…

**Exercise 2**

- State whether the following statements are true or false: (1) 7/5 is a zero of…
- Find the zeros and number of zeros of p(x) = x^2 + 9x + 18. Show them on a graph.…
- Find the zeros, the sum and the product of zeros of p(x) = 4x^2 + 12x + 5.…
- —4 and 9 are the sum and product of the zeros respectively of a quadratic…
- Find q(x) and r(x), for the quadratic polynomial p(x) = 11x — 21 + 2x^2 when…
- Divide 2x^3 + 3x^2 — 11x — 6 by x^2 + x — 6
- 4 is a zero of the cubic polynomial p(x) = x^3 — 3x^2 — 6x + 8. Find the…
- The product of two polynomials is 3x^4 + 5x^3 — 21x^2 — 53x — 30. If one of them…
- 2 + √3 and 2 — √3 are the zeros of p(x) = x^4 — 6x^3 — 26x^2 + 138x — 35. Find…
- The linear polynomial p(x) = 7x — 3 has the zero _____A. 7/3 B. 3/7 C. - 7/3…
- The cubic polynomial p(x) = x^3 — x has _____ zeros.A. 0 B. 1 C. 2 D. 3…
- The graph of p(x) = 3x — 2 — x^2 intersects the X-axis in _____ points.A. 0 B.…
- The sum of the zeros of 3x^2 + 5x — 2 is _____A. 3/5 B. - 3/5 C. 5/3 D. - 5/3…
- The graph of p(x) = 3x + 5 represents _____A. a straight line B. parabola open…
- A quadratic polynomial has no zero. Its graph _____A. touches X-axis at any…
- For the graph in figure 2.11 y = p(x) has _____ zeros. A. 1 B. 2 C. 3 D. 4…
- The product of the zeros of x^2 — 4x + 3 is _____A. 1 B. 3 C. 4 D. —4…
- a = 3, b = 5, c = 7, d = 11 in the standard notation gives the cubic…

###### Exercise 2.1

**Question 1.**Identify the type of the following polynomials: (on base of power)

(1) p(x) = x^{2} — 5x + 6

(2) p(x) = x^{2} — x^{3} + x + 1

(3) p(x) = 5x^{2} + 8x + 3

(4) p(x) = x^{3}

**Answer:**(1) Here p (x) = x^{2} — 5x + 6

Degree of the polynomial is 2. Hence it is a quadratic polynomial in x.

(2) Here p(x) = x^{2} — x^{3} + x + 1

Degree of the polynomial is 3. Hence it is a cubic polynomial in x.

(3) Here p(x) = 5x^{2} + 8x + 3

Degree of the polynomial is 2. Hence it is a quadratic polynomial in x.

(4) Here p(x) = x^{3}

Degree of the polynomial is 3. Hence it is a cubic polynomial in x.

**Question 2.**Obtain the degree of the following polynomials:

(1) p(x) = 3x —x^{4} + x^{2} + 2x^{3} + 7

(2) p(x) = x^{3} — 3x —x^{2} + 6

(3) p(x) = 3x — 9

(4) p(x) = 2x^{2} — x + 1

**Answer:**(1) Here p(x) = 3x —x^{4} + x^{2} + 2x^{3} + 7

Highest power of x in the above expression is 4. Hence, the degree of the polynomial is 4.

(2) Here p(x) = x^{3} — 3x —x^{2} + 6

Highest power of x in the above expression is 3. Hence, the degree of the polynomial is 3.

(3) Here p(x) = 3x — 9

Highest power of x in the above expression is 1. Hence, the degree of the polynomial is 1.

(4) Here p(x) = 2x^{2} — x + 1

(5) Highest power of x in the above expression is 2. Hence, the degree of the polynomial is 2.

**Question 3.**Find the coefficients of the underlined terms:

(1) p(x) = __10x__^{3}+ 7 x^{2} — 3x + 5

(2) p(x) = 7 — __5x__^{5}+ 3x^{4} + x^{2} — x

(3) p(x) = 25 — __125x__

(4) p(x) = x^{3} — __x__^{2} + x + 7

**Answer:**(1) p(x) = __10x__^{3}+ 7 x^{2} — 3x + 5

Coefficient of the underlined term is 10.

(2) p(x) = 7 — __5x__^{5}+ 3x^{4} + x^{2} — x

Coefficient of the underlined term is –5.

(3) p(x) = 25 — __125x__

Coefficient of the underlined term is –125.

(4) p(x) = x^{3} — __x__^{2} + x + 7

Coefficient of the underlined term x is **–**1.

**Question 4.**Obtain the value of the following polynomials at the given values of x:

(1) p(x) = 2x^{3} + 3x^{2} + 7x + 9 ; at x = 0, 1

(2) p(x) = 3x^{2} + 10x + 7 ; at x = —3, 1

(3) p(x) = x^{2} — 2x + 5 ; at x = —1, 5

(4) p(x) = 2x^{4} — 3x^{3} + 7x + 5 ; at x = —2, 2

**Answer:**(1) Given, p(x) = 2x^{3} + 3x^{2} + 7x + 9

p(0) = 2(0)^{3} + 3(0)^{2} + 7(0) + 9

p(0) = 0 + 0 + 9 = 9

p(1) = 2(1)^{3} + 3(1)^{2} + 7(1) + 9

p(1) = 2 + 3 + 7 + 9 = 21

(2) Given, p(x) = 3x^{2} + 10x + 7

p(–3) = 3(–3)^{2} + 10(–3) + 7

p(–3) = 27 – 30 + 7 = 4

p(1) = 3(1)^{2} + 10(1) + 7

p(1) = 3 + 10 + 7 = 20

(3) Given, p(x) = x^{2} — 2x + 5

p(–1) = (–1)^{2} – 2(–1) + 5

p(–1) = 1 + 2 + 5 = 8

p(5) = (5)^{2} – 2(5) + 5

p(5) = 25 – 10 + 5 = 20

(4) Given, p(x) = 2x^{4} — 3x^{3} + 7x + 5

p(–2) = 2(–2)^{4} – 3(–2)^{3} + 7(–2) + 5

p(–2) = 32 + 24 – 14 + 5 = 47

p(2) = 2(2)^{4} – 3(2)^{3} + 7(2) + 5

p(2) = 32– 24 + 14 + 5 = 27

**Question 5.**Examine the validity of the following statements:

(x + 1) is a factor of p(x) = 3x^{3} + 2x^{2} + 7 x + 8

**Answer:**Given, p(x) = 3x^{3} + 2x^{2} + 7 x + 8

For (x + 1) to be a factor of p(x), the condition: p(–1) = 0; should be satisfied.

p(–1) = 3(–1)^{3} + 2(–1)^{2} + 7(–1) + 8

= 3(–1) + 2(1) + 7(–1) + 8

= –3 + 2 – 7 + 8

= 0

Thus, (x + 1) is a factor of the given polynomial.

⇒ The given statement is valid.

**Question 6.**Examine the validity of the following statements:

(x + 2) is a factor of p(x) = x^{3} + x^{2} + x + 2

**Answer:**Given, p(x) = x^{3} + x^{2} + x + 2

For (x + 2) to be a factor of p(x), the condition: p(–2) = 0; should be satisfied.

p(–2) = (–2)^{3} + (–2)^{2} + (–2) + 2

= (–8) + (4) + (–2) + 2

= – 8 + 4 – 2 + 2

= –4 ≠ 0

Thus, (x + 2) is not a factor of the given polynomial.

⇒ The given statement is invalid.

**Question 7.**Examine the validity of the following statements:

(x — 1) is a factor of p(x) = x^{4} — 2x^{3} + 3x —2

**Answer:**Given, p(x) = x^{4} — 2x^{3} + 3x —2

For (x – 1) to be a factor of p(x), the condition: p(1) = 0; should be satisfied.

p(1) = (1)^{4} – 2(1)^{3} + 3(1) – 2

= 1 – 2 + 3 – 2

= 0

Thus, (x – 1) is a factor of the given polynomial.

⇒ The given statement is valid.

**Question 8.**Examine the validity of the following statements:

(x — 3) is a factor of p(x) = x^{2} — 2x — 3

**Answer:**Given, p(x) = x^{2} — 2x — 3

For (x – 3) to be a factor of p(x), the condition: p(3) = 0; should be satisfied.

p(3) = (3)^{2} – 2(3) – 3

= 9 – 6 – 3

= 0

Thus, (x –3) is a factor of the given polynomial.

⇒ The given statement is valid.

**Question 9.**Factorize the following polynomials:

p(x) = x^{3} —x^{2}—x + 1

**Answer:**p(x) = x^{3} – x^{2} – x + 1

= x^{2} (x – 1) – 1(x – 1)

= (x – 1)(x^{2} – 1)

= (x – 1)(x – 1)(x + 1)

Using the identity: (a^{2} – b^{2}) = (a – b) (a + b)

= (x – 1)^{2} (x + 1)

**Question 10.**Factorize the following polynomials:

p(x) = 5x^{2} + 11x + 6

**Answer:**p(x) = 5x^{2} + 11x + 6

= 5x^{2} + 5x + 6x + 6

(Splitting 11x as 5x + 6x)

= 5x (x + 1) + 6(x + 1)

= (x + 1) (5x + 6)

**Question 11.**Factorize the following polynomials:

p(x) = x^{3} — 3x^{2} + 9x — 27

**Answer:**p(x) = x^{3} – 3x^{2} + 9x – 27

= x^{2} (x – 3) + 9 (x – 3)

= (x – 3) (x^{2} + 9)

**Question 12.**Factorize the following polynomials:

p(x) = x^{3} + 2x^{2} + 3x + 2

**Answer:**p(x) = x^{3} + 2x^{2} + 3x + 2

= x^{3} + x^{2} + x^{2} + x + 2x + 2 (2x^{2} = x^{2} + x^{2} and 3x = x + 2x)

= x^{2} (x + 1) + x (x + 1) + 2 (x + 1)

= (x + 1) (x^{2} + x + 2)

**Question 13.**Prove that x — 2 is a factor of p(x) = x^{3} — 2x^{2}

**Answer:**Given, p(x) = x^{3} — 2x^{2}

For (x – 2) to be a factor of p(x), the condition: p(2) = 0 ;should be satisfied.

p(2) = (2)^{3} – 2(2)^{2}

= 8 – 2(4)

= 8 – 8

= 0

Hence proved that, x — 2 is a factor of p(x) = x^{3} — 2x^{2}

**Question 1.**

Identify the type of the following polynomials: (on base of power)

(1) p(x) = x^{2} — 5x + 6

(2) p(x) = x^{2} — x^{3} + x + 1

(3) p(x) = 5x^{2} + 8x + 3

(4) p(x) = x^{3}

**Answer:**

(1) Here p (x) = x^{2} — 5x + 6

Degree of the polynomial is 2. Hence it is a quadratic polynomial in x.

(2) Here p(x) = x^{2} — x^{3} + x + 1

Degree of the polynomial is 3. Hence it is a cubic polynomial in x.

(3) Here p(x) = 5x^{2} + 8x + 3

Degree of the polynomial is 2. Hence it is a quadratic polynomial in x.

(4) Here p(x) = x^{3}

Degree of the polynomial is 3. Hence it is a cubic polynomial in x.

**Question 2.**

Obtain the degree of the following polynomials:

(1) p(x) = 3x —x^{4} + x^{2} + 2x^{3} + 7

(2) p(x) = x^{3} — 3x —x^{2} + 6

(3) p(x) = 3x — 9

(4) p(x) = 2x^{2} — x + 1

**Answer:**

(1) Here p(x) = 3x —x^{4} + x^{2} + 2x^{3} + 7

Highest power of x in the above expression is 4. Hence, the degree of the polynomial is 4.

(2) Here p(x) = x^{3} — 3x —x^{2} + 6

Highest power of x in the above expression is 3. Hence, the degree of the polynomial is 3.

(3) Here p(x) = 3x — 9

Highest power of x in the above expression is 1. Hence, the degree of the polynomial is 1.

(4) Here p(x) = 2x^{2} — x + 1

(5) Highest power of x in the above expression is 2. Hence, the degree of the polynomial is 2.

**Question 3.**

Find the coefficients of the underlined terms:

(1) p(x) = __10x ^{3}__+ 7 x

^{2}— 3x + 5

(2) p(x) = 7 —

__5x__+ 3x

^{5}^{4}+ x

^{2}— x

(3) p(x) = 25 —

__125x__

(4) p(x) = x

^{3}—

__x__+ x + 7

^{2}**Answer:**

(1) p(x) = __10x ^{3}__+ 7 x

^{2}— 3x + 5

Coefficient of the underlined term is 10.

(2) p(x) = 7 — __5x ^{5}__+ 3x

^{4}+ x

^{2}— x

Coefficient of the underlined term is –5.

(3) p(x) = 25 — __125x__

Coefficient of the underlined term is –125.

(4) p(x) = x^{3} — __x ^{2}__ + x + 7

Coefficient of the underlined term x is **–**1.

**Question 4.**

Obtain the value of the following polynomials at the given values of x:

(1) p(x) = 2x^{3} + 3x^{2} + 7x + 9 ; at x = 0, 1

(2) p(x) = 3x^{2} + 10x + 7 ; at x = —3, 1

(3) p(x) = x^{2} — 2x + 5 ; at x = —1, 5

(4) p(x) = 2x^{4} — 3x^{3} + 7x + 5 ; at x = —2, 2

**Answer:**

(1) Given, p(x) = 2x^{3} + 3x^{2} + 7x + 9

p(0) = 2(0)^{3} + 3(0)^{2} + 7(0) + 9

p(0) = 0 + 0 + 9 = 9

p(1) = 2(1)^{3} + 3(1)^{2} + 7(1) + 9

p(1) = 2 + 3 + 7 + 9 = 21

(2) Given, p(x) = 3x^{2} + 10x + 7

p(–3) = 3(–3)^{2} + 10(–3) + 7

p(–3) = 27 – 30 + 7 = 4

p(1) = 3(1)^{2} + 10(1) + 7

p(1) = 3 + 10 + 7 = 20

(3) Given, p(x) = x^{2} — 2x + 5

p(–1) = (–1)^{2} – 2(–1) + 5

p(–1) = 1 + 2 + 5 = 8

p(5) = (5)^{2} – 2(5) + 5

p(5) = 25 – 10 + 5 = 20

(4) Given, p(x) = 2x^{4} — 3x^{3} + 7x + 5

p(–2) = 2(–2)^{4} – 3(–2)^{3} + 7(–2) + 5

p(–2) = 32 + 24 – 14 + 5 = 47

p(2) = 2(2)^{4} – 3(2)^{3} + 7(2) + 5

p(2) = 32– 24 + 14 + 5 = 27

**Question 5.**

Examine the validity of the following statements:

(x + 1) is a factor of p(x) = 3x^{3} + 2x^{2} + 7 x + 8

**Answer:**

Given, p(x) = 3x^{3} + 2x^{2} + 7 x + 8

For (x + 1) to be a factor of p(x), the condition: p(–1) = 0; should be satisfied.

p(–1) = 3(–1)^{3} + 2(–1)^{2} + 7(–1) + 8

= 3(–1) + 2(1) + 7(–1) + 8

= –3 + 2 – 7 + 8

= 0

Thus, (x + 1) is a factor of the given polynomial.

⇒ The given statement is valid.

**Question 6.**

Examine the validity of the following statements:

(x + 2) is a factor of p(x) = x^{3} + x^{2} + x + 2

**Answer:**

Given, p(x) = x^{3} + x^{2} + x + 2

For (x + 2) to be a factor of p(x), the condition: p(–2) = 0; should be satisfied.

p(–2) = (–2)^{3} + (–2)^{2} + (–2) + 2

= (–8) + (4) + (–2) + 2

= – 8 + 4 – 2 + 2

= –4 ≠ 0

Thus, (x + 2) is not a factor of the given polynomial.

⇒ The given statement is invalid.

**Question 7.**

Examine the validity of the following statements:

(x — 1) is a factor of p(x) = x^{4} — 2x^{3} + 3x —2

**Answer:**

Given, p(x) = x^{4} — 2x^{3} + 3x —2

For (x – 1) to be a factor of p(x), the condition: p(1) = 0; should be satisfied.

p(1) = (1)^{4} – 2(1)^{3} + 3(1) – 2

= 1 – 2 + 3 – 2

= 0

Thus, (x – 1) is a factor of the given polynomial.

⇒ The given statement is valid.

**Question 8.**

Examine the validity of the following statements:

(x — 3) is a factor of p(x) = x^{2} — 2x — 3

**Answer:**

Given, p(x) = x^{2} — 2x — 3

For (x – 3) to be a factor of p(x), the condition: p(3) = 0; should be satisfied.

p(3) = (3)^{2} – 2(3) – 3

= 9 – 6 – 3

= 0

Thus, (x –3) is a factor of the given polynomial.

⇒ The given statement is valid.

**Question 9.**

Factorize the following polynomials:

p(x) = x^{3} —x^{2}—x + 1

**Answer:**

p(x) = x^{3} – x^{2} – x + 1

= x^{2} (x – 1) – 1(x – 1)

= (x – 1)(x^{2} – 1)

= (x – 1)(x – 1)(x + 1)

Using the identity: (a^{2} – b^{2}) = (a – b) (a + b)

= (x – 1)^{2} (x + 1)

**Question 10.**

Factorize the following polynomials:

p(x) = 5x^{2} + 11x + 6

**Answer:**

p(x) = 5x^{2} + 11x + 6

= 5x^{2} + 5x + 6x + 6

(Splitting 11x as 5x + 6x)

= 5x (x + 1) + 6(x + 1)

= (x + 1) (5x + 6)

**Question 11.**

Factorize the following polynomials:

p(x) = x^{3} — 3x^{2} + 9x — 27

**Answer:**

p(x) = x^{3} – 3x^{2} + 9x – 27

= x^{2} (x – 3) + 9 (x – 3)

= (x – 3) (x^{2} + 9)

**Question 12.**

Factorize the following polynomials:

p(x) = x^{3} + 2x^{2} + 3x + 2

**Answer:**

p(x) = x^{3} + 2x^{2} + 3x + 2

= x^{3} + x^{2} + x^{2} + x + 2x + 2 (2x^{2} = x^{2} + x^{2} and 3x = x + 2x)

= x^{2} (x + 1) + x (x + 1) + 2 (x + 1)

= (x + 1) (x^{2} + x + 2)

**Question 13.**

Prove that x — 2 is a factor of p(x) = x^{3} — 2x^{2}

**Answer:**

Given, p(x) = x^{3} — 2x^{2}

For (x – 2) to be a factor of p(x), the condition: p(2) = 0 ;should be satisfied.

p(2) = (2)^{3} – 2(2)^{2}

= 8 – 2(4)

= 8 – 8

= 0

Hence proved that, x — 2 is a factor of p(x) = x^{3} — 2x^{2}

###### Exercise 2.2

**Question 1.**Find the number of zeros of the following polynomials:

p(x) = x^{2} — x

**Answer:**Here, p(x) = x^{2} — x

To find the zeros of p(x), consider p(x) = 0

∴ x^{2} – x = 0

∴ x (x – 1) = 0

∴ x = 0 or x = 1

∴ 0 and 1 are zeros of p(x).

Hence, the number of zeros of the given polynomial is 2.

**Question 2.**Find the number of zeros of the following polynomials:

p(x) = x —x^{2} —1

**Answer:**Here, p(x) = x —x^{2} —1

To find the zeros of p(x), consider p(x) = 0

∴ x – x^{2} – 1 = 0

Taking negative common

x^{2} – x + 1 = 0

x^{2} – x + + 1 – = 0

(Adding and subtracting in order to do factorization)

+ = 0

Using the identity: (a– b) ^{2} = (a^{2} – 2ab + b^{2})

= –

The above equation cannot be true as the square of a real number cannot be negative.

⇒ No zero exists for p(x).

Hence, the number of zeros of the given polynomial is 0.

**Question 3.**Find the number of zeros of the following polynomials:

P(x) = 3x —2

**Answer:**Here, p(x) = 3x —2

To find the zeros of p(x), consider p(x) = 0

∴ 3x – 2 = 0

∴ x =

∴ is a zero of p(x).

Hence, the number of zeros of the given polynomial is 1.

**Question 4.**Find the number of zeros of the following polynomials:

p(x) = x^{3} — x

**Answer:**Here, p(x) = x^{3} — x

To find the zeros of p(x), consider p(x) = 0

∴ x^{3} – x = 0

∴ x (x^{2} – 1) = 0

∴ x (x – 1) (x + 1) = 0 Using the identity: (a^{2} – b^{2}) = (a – b) (a + b)

∴ x = 0 or x = –1 or x = 1

∴ 0, –1 and 1 are the zeros of p(x).

Thus, the number of zeros of the given polynomial is 3.

**Question 5.**Find the number of zeros and real zeros of p(x) = x^{3} + 1. Show them by a graph.

**Answer:**Given, p(x) = x^{3} + 1

To find the zeros of p(x), consider p(x) = 0

∴ x^{3} + 1 = 0

∴ (x + 1) (x^{2} – x + 1) = 0 Using the identity (a + b)^{3} = (a + b) (a^{2} – ab + b^{2})

∴ x + 1 = 0 or x^{2} – x + 1 = 0

∴ x = –1 ; real zeros of x^{2} – x + 1 are not possible.

∴ –1 is the only zero of p(x).

Since –1 is real, number of real zeros is 1.

The given polynomial has degree 3, so it can have 3 zeros at most.

⇒ No. of zeros = 3 ; No. of real zeros = 1

For p(x) = x^{3} + 1, taking x = –2, –1, 0 and 2.

We obtain the following table:

Plotting these points on graph paper, we obtain the following figure:

It can be seen that the graph intersects the X-axis at (–1, 0), so –1 is the zero of p(x).

**Question 6.**Draw the graph of p(x) = x^{2} + 1 and find the real zeros of this polynomial.

**Answer:**Taking values as x = –3, –2, –1, 0, 1, 2, 3 in p(x), we obtain the following table:

Plotting these points on graph paper, we obtain the following figure:

The graph does not intersect the X -axis at any point.

∴ p(x) has no real zero.

**Question 7.**From the figure 2.10 find the number of zeros of y = p(x).

(i) (ii)

(iii) (iv)

(v) (vi)

Figure 2.10

**Answer:**(i) The graph of y = p(x) intersects X-axis at one point. So, the number of real zeros of p(x) is one.

(ii) The graph of y = p(x) does not intersect X-axis at any point. So, p(x) has no real zeros.

(iii) The graph of y = p(x) intersects X-axis at three points. So, the number of real zeros of p(x) is three.

(iv) The graph of y = p(x) intersects X-axis at two points. So, the number of real zeros of p(x) is two.

(v) The graph of y = p(x) intersects X-axis at four points. So, the number of real zeros of p(x) is four.

(vi) The graph of y = p(x) intersects X-axis at three points. So, the number of real zeros of p(x) is three.

**Question 8.**Find the number of zeros and zeros of p(x) = x^{2} — 4. Represent them graphically.

**Answer:**Given, p(x) = x^{2} – 4

To find the zeros of p(x), consider p(x) = 0

∴ x^{2} – 4 = 0

(x – 2)(x + 2) = 0

Using the identity: (a^{2} – b^{2}) = (a – b) (a + b)

x = 2 or x = –2

⇒ The real zeros of p(x) are 2 and –2, i.e. p(x) has two zeros.

To draw the graph of this polynomial, consider the following values of x and corresponding values of p(x):

Plotting these points on a graph paper as shown in the figure:

It can be observed that the graph of p(x) intersects the X-axis at two distinct points (–2, 0) and (2, 0).

The X–coordinates of these point are considered as zeros of p(x).

Thus, –2 and 2 are the zeros of p(x).

**Question 1.**

Find the number of zeros of the following polynomials:

p(x) = x^{2} — x

**Answer:**

Here, p(x) = x^{2} — x

To find the zeros of p(x), consider p(x) = 0

∴ x^{2} – x = 0

∴ x (x – 1) = 0

∴ x = 0 or x = 1

∴ 0 and 1 are zeros of p(x).

Hence, the number of zeros of the given polynomial is 2.

**Question 2.**

Find the number of zeros of the following polynomials:

p(x) = x —x^{2} —1

**Answer:**

Here, p(x) = x —x^{2} —1

To find the zeros of p(x), consider p(x) = 0

∴ x – x^{2} – 1 = 0

Taking negative common

x^{2} – x + 1 = 0

x^{2} – x + + 1 – = 0

(Adding and subtracting in order to do factorization)

+ = 0

Using the identity: (a– b) ^{2} = (a^{2} – 2ab + b^{2})

= –

The above equation cannot be true as the square of a real number cannot be negative.

⇒ No zero exists for p(x).

Hence, the number of zeros of the given polynomial is 0.

**Question 3.**

Find the number of zeros of the following polynomials:

P(x) = 3x —2

**Answer:**

Here, p(x) = 3x —2

To find the zeros of p(x), consider p(x) = 0

∴ 3x – 2 = 0

∴ x =

∴ is a zero of p(x).

Hence, the number of zeros of the given polynomial is 1.

**Question 4.**

Find the number of zeros of the following polynomials:

p(x) = x^{3} — x

**Answer:**

Here, p(x) = x^{3} — x

To find the zeros of p(x), consider p(x) = 0

∴ x^{3} – x = 0

∴ x (x^{2} – 1) = 0

∴ x (x – 1) (x + 1) = 0 Using the identity: (a^{2} – b^{2}) = (a – b) (a + b)

∴ x = 0 or x = –1 or x = 1

∴ 0, –1 and 1 are the zeros of p(x).

Thus, the number of zeros of the given polynomial is 3.

**Question 5.**

Find the number of zeros and real zeros of p(x) = x^{3} + 1. Show them by a graph.

**Answer:**

Given, p(x) = x^{3} + 1

To find the zeros of p(x), consider p(x) = 0

∴ x^{3} + 1 = 0

∴ (x + 1) (x^{2} – x + 1) = 0 Using the identity (a + b)^{3} = (a + b) (a^{2} – ab + b^{2})

∴ x + 1 = 0 or x^{2} – x + 1 = 0

∴ x = –1 ; real zeros of x^{2} – x + 1 are not possible.

∴ –1 is the only zero of p(x).

Since –1 is real, number of real zeros is 1.

The given polynomial has degree 3, so it can have 3 zeros at most.

⇒ No. of zeros = 3 ; No. of real zeros = 1

For p(x) = x^{3} + 1, taking x = –2, –1, 0 and 2.

We obtain the following table:

Plotting these points on graph paper, we obtain the following figure:

It can be seen that the graph intersects the X-axis at (–1, 0), so –1 is the zero of p(x).

**Question 6.**

Draw the graph of p(x) = x^{2} + 1 and find the real zeros of this polynomial.

**Answer:**

Taking values as x = –3, –2, –1, 0, 1, 2, 3 in p(x), we obtain the following table:

Plotting these points on graph paper, we obtain the following figure:

The graph does not intersect the X -axis at any point.

∴ p(x) has no real zero.

**Question 7.**

From the figure 2.10 find the number of zeros of y = p(x).

(i) (ii)

(iii) (iv)

(v) (vi)

Figure 2.10

**Answer:**

(i) The graph of y = p(x) intersects X-axis at one point. So, the number of real zeros of p(x) is one.

(ii) The graph of y = p(x) does not intersect X-axis at any point. So, p(x) has no real zeros.

(iii) The graph of y = p(x) intersects X-axis at three points. So, the number of real zeros of p(x) is three.

(iv) The graph of y = p(x) intersects X-axis at two points. So, the number of real zeros of p(x) is two.

(v) The graph of y = p(x) intersects X-axis at four points. So, the number of real zeros of p(x) is four.

(vi) The graph of y = p(x) intersects X-axis at three points. So, the number of real zeros of p(x) is three.

**Question 8.**

Find the number of zeros and zeros of p(x) = x^{2} — 4. Represent them graphically.

**Answer:**

Given, p(x) = x^{2} – 4

To find the zeros of p(x), consider p(x) = 0

∴ x^{2} – 4 = 0

(x – 2)(x + 2) = 0

Using the identity: (a^{2} – b^{2}) = (a – b) (a + b)

x = 2 or x = –2

⇒ The real zeros of p(x) are 2 and –2, i.e. p(x) has two zeros.

To draw the graph of this polynomial, consider the following values of x and corresponding values of p(x):

Plotting these points on a graph paper as shown in the figure:

It can be observed that the graph of p(x) intersects the X-axis at two distinct points (–2, 0) and (2, 0).

The X–coordinates of these point are considered as zeros of p(x).

Thus, –2 and 2 are the zeros of p(x).

###### Exercise 2.3

**Question 1.**Prove that 4 and 1 are the zeros of the quadratic polynomial p(x) = x^{2} — 5x + 4. Also verify the relationship between the zeros and the coefficients of p(x).

**Answer:**Here p(x) = x^{2} – 5x + 4

To prove that 4 and 1 are the zeros of p(x), we need to prove that p(4) = p(1) = 0.

p(4) = (4)^{2} – 5(4) + 4

= 16 – 20 + 4

= 0

p(1) = (1)^{2} – 5(1) + 4

= 1– 5 + 4

= 0

Hence proved that 4 and 1 are the zeros of the quadratic polynomial p(x) = x^{2} – 5x + 4.

In p(x) = x^{2} – 5x + 4

a = 1, b = –5, c = 4

Sum of zeros = 4 + 1 = 5

= – (–5)

= – = – = –

Product of zeros = 4×1 = 4

= = =

Hence verified the relationship between the zeros and the coefficients of p(x).

**Question 2.**Find the zeros of the following quadratic polynomials:

p(x) = x^{2} + 4x — 21

**Answer:**Here p(x) = x^{2} + 4x – 21

To find the zeros of p(x), let p(x) = 0

∴ x^{2} + 4x – 21 = 0

∴ x^{2} + 7x – 3x – 21 = 0

∴ x(x + 7) – 3(x + 7) = 0

∴ (x + 7)(x – 3) = 0

∴ x = –7 or x = 3

⇒ –7 and 3 are the zeros of the quadratic polynomial p(x).

**Question 3.**Find the zeros of the following quadratic polynomials:

p(x) = 6x^{2} — 11x + 5

**Answer:**Here p(x) = 6x^{2} — 11x + 5

To find the zeros of p(x), let p(x) = 0

∴6x^{2} — 11x + 5 = 0

∴ 6x^{2} – 6x – 5x + 5 = 0

∴ 6x(x – 1) – 5(x – 5) = 0

∴ (x – 1)(6x – 5) = 0

∴ x = 1 or x =

⇒ 1 and are the zeros of the quadratic polynomial p(x).

**Question 4.**Find the zeros of the following quadratic polynomials:

p(x) = 4x^{2} + 9x + 5

**Answer:**Here p(x) = 4x^{2} + 9x + 5

To find the zeros of p(x), let p(x) = 0

∴4x^{2} + 9x + 5 = 0

∴4x^{2} + 4x + 5x + 5 = 0

∴4x(x + 1) + 5(x + 1) = 0

∴ (x + 1)(4x + 5) = 0

∴ x = –1 or x = –

⇒ 1 and – are the zeros of the quadratic polynomial p(x).

**Question 5.**Find the zeros of the following quadratic polynomials:

p(x) = 3x^{2} + 5x — 8

**Answer:**Here p(x) = 3x^{2} + 5x — 8

To find the zeros of p(x), let p(x) = 0

∴3x^{2} + 5x — 8 = 0

∴ 3x^{2} + 8x – 3x – 8 = 0

∴ x(3x + 8) – 1(3x + 8) = 0

∴ (3x + 8)(x – 1) = 0

∴ x = 1 or x = –

⇒ 1 and – are the zeros of the quadratic polynomial p(x).

**Question 6.**Find the zeros of the following quadratic polynomials:

p(x) = x^{2} — 81

**Answer:**Here p(x) = x^{2} — 81

To find the zeros of p(x), let p(x) = 0

∴ x^{2} – 81 = 0

∴ (x)^{2} – (9)^{2} = 0

∴ (x – 9) (x + 9) = 0

[Using the identity: (a^{2} – b^{2}) = (a – b) (a + b)]

∴ x = 9 or x = –9

–9 and 9 are the zeros of the quadratic polynomial p(x).

**Question 7.**Find the zeros of the following quadratic polynomials:

p(x) = x^{2} — x — 6

**Answer:**Here p(x) = x^{2} — x — 6

To find the zeros of p(x), let p(x) = 0

∴ x^{2} — x — 6 = 0

∴ x^{2} — 3x + 2x — 6 = 0

∴ x(x – 3) + 2 (x – 3) = 0

∴ (x – 3) (x + 2) = 0

∴ x = 3 or x = –2

3 and –2 are the zeros of the quadratic polynomial p(x).

**Question 8.**Find the zeros, the sum of the zeros and the product of the zeros of the quadratic polynomial p(x) = 3x^{2} — x — 4

**Answer:**Here p(x) = 3x^{2} — x — 4

To find the zeros of p(x), let p(x) = 0

∴3x^{2} — x — 4 = 0

∴ 3x^{2} + 3x – 4x – 4 = 0

∴ 3x(x + 1) – 4(x + 1) = 0

∴ (3x – 4)(x + 1) = 0

∴ x = or x = –1

⇒ and –1are the zeros of the quadratic polynomial p(x).

Sum of zeros = + (–1) =

Product of zeros = × (–1) = –

**Question 9.**Obtain a quadratic polynomial with the following conditions:

The sum of zeros = 2; the product of zeros = —3

**Answer:**Let Î± and Î² be the zeros of the polynomial p(x) = ax^{2} + bx + c.

Given, Î± + Î² = 2 ; Î±Î² = –3

We know that **Î± + Î² = –** = – . So, – = = k, say

Thus, b = –2k, a = k

And **Î±Î² =** = – . So, c = –3a = –3(k) = –3k

p(x) = ax^{2} + bx + c

∴ p(x) = (k)x^{2} + (–2k)x – 3k

∴ p(x) = k (x^{2} – 2x – 3)

**Question 10.**Obtain a quadratic polynomial with the following conditions:

The sum of zeros = —3; the product of zeros = —4

**Answer:**Let Î± and Î² be the zeros of the polynomial p(x) = ax^{2} + bx + c.

Given, Î± + Î² = –3 ; Î±Î² = –4

We know that **Î± + Î² = –** = – . So, – = = k, say

Thus, b = –3k, a = k

And **Î±Î² =** = – . So, c = –4a = –4(k) = –4k

p(x) = ax^{2} + bx + c

∴ p(x) = (k)x^{2} + (–3k)x – 4k

∴ p(x) = k (x^{2} – 3x– 4)

**Question 11.**Obtain a quadratic polynomial with the following conditions:

The sum of zeros = 1/3; the product of zeros = 1/2

**Answer:**Let Î± and Î² be the zeros of the polynomial p(x) = ax^{2} + bx + c.

Given, Î± + Î² = ; Î±Î² =

We know that **Î± + Î² = –** = . So, – = = k, say

Thus, b = –k, a = 3k

And **Î±Î² =** = . So, c = =

p(x) = ax^{2} + bx + c

∴ p(x) = (3k)x^{2} + (–k)x +

∴ p(x) = k (3x^{2} – x + )

∴ p(x) = (6x^{2} – 2x + 3) [Taking common]

**Question 12.**Obtain the quadratic or the cubic polynomial as the case may be in the standard form with the following coefficients:

(1) a = 6, b = 17, c = 11

(2) a = 1, b = —1, c = —1, d = 1

(3) a = 5, b = 7, c = 2

(4) a = 1, b = —3, c = —1, d = 3

(5) a = 3, b = —5, c = —11, d = —3

**Answer:**(1) The quadratic polynomial is,

p(x) = ax^{2} + bx + c; a ≠ 0, a, b, c Îµ R.

∴ Substituting a = 6, b = 17 and c = 11, we get the required quadratic polynomial

p(x) = 6x^{2} + 17x + 11.

(2) The cubic polynomial is,

p(x) = ax^{3} + bx^{2} + cx + d; a ≠ 0, a, b, c, d Îµ R.

∴ Substituting a = 1, b = —1, c = —1 and d = 1, we get the required cubic polynomial

p(x) = x^{3} – x^{2} – x + 1.

(3) The quadratic polynomial is,

p(x) = ax + bx + c; a ≠ 0, a, b, c Îµ R.

∴ Substituting a = 5, b = 7 and c = 2, we get the required quadratic polynomial

p(x) = 5x^{2} + 7x + 2.

(4) The cubic polynomial is,

p(x) = ax + bx + cx + d; a ≠ 0, a, b, c, d Îµ R.

∴ Substituting a = 1, b = –3, c = –1 and d = 3, we get the required cubic polynomial

p(x) = x^{3} – 3x^{2} – x + 3.

(5) The cubic polynomial is,

p(x) = ax^{3} + bx^{2} + cx + d; a ≠ 0, a, b, c, d Îµ R.

∴ Substituting a = 3, b = –5, c = –11 and d = –3, we get the required cubic polynomial

p(x) = 3x^{3} – 5x^{2} – 11x – 3.

**Question 1.**

Prove that 4 and 1 are the zeros of the quadratic polynomial p(x) = x^{2} — 5x + 4. Also verify the relationship between the zeros and the coefficients of p(x).

**Answer:**

Here p(x) = x^{2} – 5x + 4

To prove that 4 and 1 are the zeros of p(x), we need to prove that p(4) = p(1) = 0.

p(4) = (4)^{2} – 5(4) + 4

= 16 – 20 + 4

= 0

p(1) = (1)^{2} – 5(1) + 4

= 1– 5 + 4

= 0

Hence proved that 4 and 1 are the zeros of the quadratic polynomial p(x) = x^{2} – 5x + 4.

In p(x) = x^{2} – 5x + 4

a = 1, b = –5, c = 4

Sum of zeros = 4 + 1 = 5

= – (–5)

= – = – = –

Product of zeros = 4×1 = 4

= = =

Hence verified the relationship between the zeros and the coefficients of p(x).

**Question 2.**

Find the zeros of the following quadratic polynomials:

p(x) = x^{2} + 4x — 21

**Answer:**

Here p(x) = x^{2} + 4x – 21

To find the zeros of p(x), let p(x) = 0

∴ x^{2} + 4x – 21 = 0

∴ x^{2} + 7x – 3x – 21 = 0

∴ x(x + 7) – 3(x + 7) = 0

∴ (x + 7)(x – 3) = 0

∴ x = –7 or x = 3

⇒ –7 and 3 are the zeros of the quadratic polynomial p(x).

**Question 3.**

Find the zeros of the following quadratic polynomials:

p(x) = 6x^{2} — 11x + 5

**Answer:**

Here p(x) = 6x^{2} — 11x + 5

To find the zeros of p(x), let p(x) = 0

∴6x^{2} — 11x + 5 = 0

∴ 6x^{2} – 6x – 5x + 5 = 0

∴ 6x(x – 1) – 5(x – 5) = 0

∴ (x – 1)(6x – 5) = 0

∴ x = 1 or x =

⇒ 1 and are the zeros of the quadratic polynomial p(x).

**Question 4.**

Find the zeros of the following quadratic polynomials:

p(x) = 4x^{2} + 9x + 5

**Answer:**

Here p(x) = 4x^{2} + 9x + 5

To find the zeros of p(x), let p(x) = 0

∴4x^{2} + 9x + 5 = 0

∴4x^{2} + 4x + 5x + 5 = 0

∴4x(x + 1) + 5(x + 1) = 0

∴ (x + 1)(4x + 5) = 0

∴ x = –1 or x = –

⇒ 1 and – are the zeros of the quadratic polynomial p(x).

**Question 5.**

Find the zeros of the following quadratic polynomials:

p(x) = 3x^{2} + 5x — 8

**Answer:**

Here p(x) = 3x^{2} + 5x — 8

To find the zeros of p(x), let p(x) = 0

∴3x^{2} + 5x — 8 = 0

∴ 3x^{2} + 8x – 3x – 8 = 0

∴ x(3x + 8) – 1(3x + 8) = 0

∴ (3x + 8)(x – 1) = 0

∴ x = 1 or x = –

⇒ 1 and – are the zeros of the quadratic polynomial p(x).

**Question 6.**

Find the zeros of the following quadratic polynomials:

p(x) = x^{2} — 81

**Answer:**

Here p(x) = x^{2} — 81

To find the zeros of p(x), let p(x) = 0

∴ x^{2} – 81 = 0

∴ (x)^{2} – (9)^{2} = 0

∴ (x – 9) (x + 9) = 0

[Using the identity: (a^{2} – b^{2}) = (a – b) (a + b)]

∴ x = 9 or x = –9

–9 and 9 are the zeros of the quadratic polynomial p(x).

**Question 7.**

Find the zeros of the following quadratic polynomials:

p(x) = x^{2} — x — 6

**Answer:**

Here p(x) = x^{2} — x — 6

To find the zeros of p(x), let p(x) = 0

∴ x^{2} — x — 6 = 0

∴ x^{2} — 3x + 2x — 6 = 0

∴ x(x – 3) + 2 (x – 3) = 0

∴ (x – 3) (x + 2) = 0

∴ x = 3 or x = –2

3 and –2 are the zeros of the quadratic polynomial p(x).

**Question 8.**

Find the zeros, the sum of the zeros and the product of the zeros of the quadratic polynomial p(x) = 3x^{2} — x — 4

**Answer:**

Here p(x) = 3x^{2} — x — 4

To find the zeros of p(x), let p(x) = 0

∴3x^{2} — x — 4 = 0

∴ 3x^{2} + 3x – 4x – 4 = 0

∴ 3x(x + 1) – 4(x + 1) = 0

∴ (3x – 4)(x + 1) = 0

∴ x = or x = –1

⇒ and –1are the zeros of the quadratic polynomial p(x).

Sum of zeros = + (–1) =

Product of zeros = × (–1) = –

**Question 9.**

Obtain a quadratic polynomial with the following conditions:

The sum of zeros = 2; the product of zeros = —3

**Answer:**

Let Î± and Î² be the zeros of the polynomial p(x) = ax^{2} + bx + c.

Given, Î± + Î² = 2 ; Î±Î² = –3

We know that **Î± + Î² = –** = – . So, – = = k, say

Thus, b = –2k, a = k

And **Î±Î² =** = – . So, c = –3a = –3(k) = –3k

p(x) = ax^{2} + bx + c

∴ p(x) = (k)x^{2} + (–2k)x – 3k

∴ p(x) = k (x^{2} – 2x – 3)

**Question 10.**

Obtain a quadratic polynomial with the following conditions:

The sum of zeros = —3; the product of zeros = —4

**Answer:**

Let Î± and Î² be the zeros of the polynomial p(x) = ax^{2} + bx + c.

Given, Î± + Î² = –3 ; Î±Î² = –4

We know that **Î± + Î² = –** = – . So, – = = k, say

Thus, b = –3k, a = k

And **Î±Î² =** = – . So, c = –4a = –4(k) = –4k

p(x) = ax^{2} + bx + c

∴ p(x) = (k)x^{2} + (–3k)x – 4k

∴ p(x) = k (x^{2} – 3x– 4)

**Question 11.**

Obtain a quadratic polynomial with the following conditions:

The sum of zeros = 1/3; the product of zeros = 1/2

**Answer:**

Let Î± and Î² be the zeros of the polynomial p(x) = ax^{2} + bx + c.

Given, Î± + Î² = ; Î±Î² =

We know that **Î± + Î² = –** = . So, – = = k, say

Thus, b = –k, a = 3k

And **Î±Î² =** = . So, c = =

p(x) = ax^{2} + bx + c

∴ p(x) = (3k)x^{2} + (–k)x +

∴ p(x) = k (3x^{2} – x + )

∴ p(x) = (6x^{2} – 2x + 3) [Taking common]

**Question 12.**

Obtain the quadratic or the cubic polynomial as the case may be in the standard form with the following coefficients:

(1) a = 6, b = 17, c = 11

(2) a = 1, b = —1, c = —1, d = 1

(3) a = 5, b = 7, c = 2

(4) a = 1, b = —3, c = —1, d = 3

(5) a = 3, b = —5, c = —11, d = —3

**Answer:**

(1) The quadratic polynomial is,

p(x) = ax^{2} + bx + c; a ≠ 0, a, b, c Îµ R.

∴ Substituting a = 6, b = 17 and c = 11, we get the required quadratic polynomial

p(x) = 6x^{2} + 17x + 11.

(2) The cubic polynomial is,

p(x) = ax^{3} + bx^{2} + cx + d; a ≠ 0, a, b, c, d Îµ R.

∴ Substituting a = 1, b = —1, c = —1 and d = 1, we get the required cubic polynomial

p(x) = x^{3} – x^{2} – x + 1.

(3) The quadratic polynomial is,

p(x) = ax + bx + c; a ≠ 0, a, b, c Îµ R.

∴ Substituting a = 5, b = 7 and c = 2, we get the required quadratic polynomial

p(x) = 5x^{2} + 7x + 2.

(4) The cubic polynomial is,

p(x) = ax + bx + cx + d; a ≠ 0, a, b, c, d Îµ R.

∴ Substituting a = 1, b = –3, c = –1 and d = 3, we get the required cubic polynomial

p(x) = x^{3} – 3x^{2} – x + 3.

(5) The cubic polynomial is,

p(x) = ax^{3} + bx^{2} + cx + d; a ≠ 0, a, b, c, d Îµ R.

∴ Substituting a = 3, b = –5, c = –11 and d = –3, we get the required cubic polynomial

p(x) = 3x^{3} – 5x^{2} – 11x – 3.

###### Exercise 2.4

**Question 1.**Divide the following polynomial p(x) by the polynomial s(x).

p(x) = 2x^{3} — 13x^{2} + 23x — 12, s(x) = 2x — 3

**Answer:**Here, dividend polynomial = p(x) = 2x^{3} – 13x^{2} + 23x – 12

and divisor polynomial = s(x) = 2x — 3

Thus, the quotient polynomial q(x) = x^{2} – 5x + 4 and the reminder polynomial r(x) = 0.

**Question 2.**Divide the following polynomial p(x) by the polynomial s(x).

p(x) = x^{2} + 5x + 6, s(x) = x + 6

**Answer:**Here, dividend polynomial = p(x) = x^{2} + 5x + 6

and divisor polynomial = s(x) = x + 6

Thus, the quotient polynomial q(x) = x + 1 and the reminder polynomial r(x) = 0.

**Question 3.**Divide the following polynomial p(x) by the polynomial s(x).

p(x) = 40x^{2} + 11x — 63, s(x) = 8x — 9

**Answer:**Here, dividend polynomial = p(x) = 40x^{2} + 11x — 63

and divisor polynomial = s(x) = 8x — 9

Thus, the quotient polynomial q(x) = 5x + 7 and the reminder polynomial r(x) = 0.

**Question 4.**Divide the following polynomial p(x) by the polynomial s(x).

p(x) = 2x^{3} + 9x^{2} + 13x + 6, s(x) = 2x^{2} + 5x + 3

**Answer:**Here, dividend polynomial = p(x) = 2x^{3} + 9x^{2} + 13x + 6

and divisor polynomial = s(x) = 2x^{2} + 5x + 3

Thus, the quotient polynomial q(x) = x + 2 and the reminder polynomial r(x) = 0.

**Question 5.**Divide the following polynomial p(x) by the polynomial s(x).

p(x) = x^{4} + 4x^{3} + 5x^{2} — 7x — 3, s(x) = x^{2} — 1

**Answer:**Here, dividend polynomial = p(x) = 2x^{3} – 13x^{2} + 23x – 12

and divisor polynomial = s(x) = 2x — 3

Thus, the quotient polynomial q(x) = x^{2} + 4x + 6 and the reminder polynomial r(x) = –3x + 3.

**Question 6.**Find the remainder polynomial when the cubic polynomial x^{3} — 3x^{2} + 4x + 5 is divided by x — 2

**Answer:**Here the dividend polynomial p(x) = x^{3} — 3x^{2} + 4x + 5 and the divisor polynomial s(x) = x – 2

Coefficients of x^{3}, x^{2}, x and x^{0} in the dividend polynomial are 1, –3, 4 and 5 respectively.

Equating divisor polynomial x – 2 to 0, we get x = 2.

The reminder obtained when the cubic polynomial x^{3} — 3x^{2} + 4x + 5 is divided by x–2 is r(x) = 9.

**Question 7.**3 is a zero of p(x) = 3x^{3} — x^{2} — ax — 45. Find 'a'.

**Answer:**Given, 3 is a zero of p(x).

⇒ p(3) = 0

∴ 3(3)^{3} — (3)^{2} — a(3) — 45 = 0

∴ 81 – 9 – 3a – 45 = 0

∴ 3a = 27

∴ a = 9

**Question 8.**The product of two polynomials is 6x^{3} + 29x^{2} + 44x + 21 and one of the polynomials is 3x + 7. Find the other polynomial.

**Answer:**Here p(x) = the dividend polynomial = 6x^{3} + 29x^{2} + 44x + 21

s(x) = the divisor polynomial = 3x + 7

Now,

∴ The quotient polynomial q(x) = 3x + 7 and the remainder polynomial r(x) = 0.

∴ The other polynomial is 2x^{2} + 5x + 3

**Question 9.**If polynomial p(x) is divided by x^{2} + 3x + 5, the quotient polynomial and the remainder polynomials are 2x^{2} + x + 1 and x — 3 respectively. Find p(x).

**Answer:**Here, divisor polynomial s(x) = x^{2} + 3x + 5

quotient polynomial q(x) = 2x^{2} + x + 1

remainder polynomial r(x) = x – 3

We know that,

Dividend = Divisor × Quotient + Remainder

∴ p(x) = s(x) × q(x) + r(x)

∴ p(x) = (x^{2} + 3x + 5) × (2x^{2} + x + 1) + (x – 3)

∴ p(x) = 2x^{4} + x^{3} + x^{2} + 6x^{3} + 3x^{2} + 3x + 10x^{2} + 5x + 5 + x – 3

∴ p(x) = 2x^{4} + 7x^{3} + 14x^{2} + 9x + 2

**Question 10.**Divide p(x) = x^{3} — 4x^{2} + 5x — 2 by x — 2 Find r(x).

**Answer:**Here the dividend polynomial p(x) = x^{3} — 4x^{2} + 5x — 2

and the divisor polynomial s(x) = x – 2

Coefficients of x^{3}, x^{2}, x and x^{0} in the dividend polynomial are 1, –4, 5 and –2 respectively.

Equating divisor polynomial x – 2 to 0, we get x = 2.

The reminder obtained when the cubic polynomial x^{3} — 4x^{2} + 5x — 2 is divided by x–2 is r(x) = 0.

**Question 11.**There are x^{4} + 57x + 15 pens to be distributed in a class of x^{2} + 4x + 2 students. Each student should get the maximum possible number of pens. Find the number of pens received by each student and the number of pens left undistributed (x N).

**Answer:**Here, dividend polynomial = p(x) = x^{4} + 57x + 15

and divisor polynomial = s(x) = x^{2} + 4x + 2

Thus, the quotient polynomial q(x) = x^{2} – 4x + 14 and the reminder polynomial r(x) = 9x – 13.

⇒ Each student will get (x^{2} – 4x + 14) pens and number of pens left undistributed = 9x – 13

**Question 12.**A trader bought 2x^{2} — x + 2 TV sets for Rs. 8x^{4} + 7x — 6. Find the price of one TV set.

**Answer:**Cost of 2x^{2} — x + 2 TV sets = Rs. 8x^{4} + 7x — 6

Here, dividend polynomial = p(x) = 8x^{4} + 7x — 6

and divisor polynomial = s(x) = 2x^{2} — x + 2

⇒ Cost of one TV set = 4x^{2} + 2x – 3.

**Question 13.** and are two of the zeros of p(x) = 2x^{4} + 7x^{3} — 8x^{2} — 14x + 8. Find the remaining zeros of p(x).

**Answer:**Given, – √2 and √2 are zeros of p(x).

⇒ (x + √2) and (x – √2) are the factors of p(x).

Thus (x + √2) (x – √2) = (x^{2} – 2) is also a factor of p(x).

To find the remaining zeros, we find the remaining factors using the division process.

Here, dividend polynomial = p(x) = 2x^{4} + 7x^{3} — 8x^{2} — 14x + 8

and divisor polynomial = s(x) = x^{2} – 2

⇒ p(x) = 2x^{4} + 7x^{3} — 8x^{2} — 14x + 8 = (x^{2} – 2)(2x^{2} + 7x — 4)

On factorising 2x^{2} + 7x — 4, we get

2x^{2} + 7x — 4 = 2x^{2} + 8x – x — 4

= 2x (x + 4) –1 (x + 4)

= (2x –1) (x + 4)

Hence the other two zeros of p(x) are and –4.

**Question 1.**

Divide the following polynomial p(x) by the polynomial s(x).

p(x) = 2x^{3} — 13x^{2} + 23x — 12, s(x) = 2x — 3

**Answer:**

Here, dividend polynomial = p(x) = 2x^{3} – 13x^{2} + 23x – 12

and divisor polynomial = s(x) = 2x — 3

Thus, the quotient polynomial q(x) = x^{2} – 5x + 4 and the reminder polynomial r(x) = 0.

**Question 2.**

Divide the following polynomial p(x) by the polynomial s(x).

p(x) = x^{2} + 5x + 6, s(x) = x + 6

**Answer:**

Here, dividend polynomial = p(x) = x^{2} + 5x + 6

and divisor polynomial = s(x) = x + 6

Thus, the quotient polynomial q(x) = x + 1 and the reminder polynomial r(x) = 0.

**Question 3.**

Divide the following polynomial p(x) by the polynomial s(x).

p(x) = 40x^{2} + 11x — 63, s(x) = 8x — 9

**Answer:**

Here, dividend polynomial = p(x) = 40x^{2} + 11x — 63

and divisor polynomial = s(x) = 8x — 9

Thus, the quotient polynomial q(x) = 5x + 7 and the reminder polynomial r(x) = 0.

**Question 4.**

Divide the following polynomial p(x) by the polynomial s(x).

p(x) = 2x^{3} + 9x^{2} + 13x + 6, s(x) = 2x^{2} + 5x + 3

**Answer:**

Here, dividend polynomial = p(x) = 2x^{3} + 9x^{2} + 13x + 6

and divisor polynomial = s(x) = 2x^{2} + 5x + 3

Thus, the quotient polynomial q(x) = x + 2 and the reminder polynomial r(x) = 0.

**Question 5.**

Divide the following polynomial p(x) by the polynomial s(x).

p(x) = x^{4} + 4x^{3} + 5x^{2} — 7x — 3, s(x) = x^{2} — 1

**Answer:**

Here, dividend polynomial = p(x) = 2x^{3} – 13x^{2} + 23x – 12

and divisor polynomial = s(x) = 2x — 3

Thus, the quotient polynomial q(x) = x^{2} + 4x + 6 and the reminder polynomial r(x) = –3x + 3.

**Question 6.**

Find the remainder polynomial when the cubic polynomial x^{3} — 3x^{2} + 4x + 5 is divided by x — 2

**Answer:**

Here the dividend polynomial p(x) = x^{3} — 3x^{2} + 4x + 5 and the divisor polynomial s(x) = x – 2

Coefficients of x^{3}, x^{2}, x and x^{0} in the dividend polynomial are 1, –3, 4 and 5 respectively.

Equating divisor polynomial x – 2 to 0, we get x = 2.

The reminder obtained when the cubic polynomial x^{3} — 3x^{2} + 4x + 5 is divided by x–2 is r(x) = 9.

**Question 7.**

3 is a zero of p(x) = 3x^{3} — x^{2} — ax — 45. Find 'a'.

**Answer:**

Given, 3 is a zero of p(x).

⇒ p(3) = 0

∴ 3(3)^{3} — (3)^{2} — a(3) — 45 = 0

∴ 81 – 9 – 3a – 45 = 0

∴ 3a = 27

∴ a = 9

**Question 8.**

The product of two polynomials is 6x^{3} + 29x^{2} + 44x + 21 and one of the polynomials is 3x + 7. Find the other polynomial.

**Answer:**

Here p(x) = the dividend polynomial = 6x^{3} + 29x^{2} + 44x + 21

s(x) = the divisor polynomial = 3x + 7

Now,

∴ The quotient polynomial q(x) = 3x + 7 and the remainder polynomial r(x) = 0.

∴ The other polynomial is 2x^{2} + 5x + 3

**Question 9.**

If polynomial p(x) is divided by x^{2} + 3x + 5, the quotient polynomial and the remainder polynomials are 2x^{2} + x + 1 and x — 3 respectively. Find p(x).

**Answer:**

Here, divisor polynomial s(x) = x^{2} + 3x + 5

quotient polynomial q(x) = 2x^{2} + x + 1

remainder polynomial r(x) = x – 3

We know that,

Dividend = Divisor × Quotient + Remainder

∴ p(x) = s(x) × q(x) + r(x)

∴ p(x) = (x^{2} + 3x + 5) × (2x^{2} + x + 1) + (x – 3)

∴ p(x) = 2x^{4} + x^{3} + x^{2} + 6x^{3} + 3x^{2} + 3x + 10x^{2} + 5x + 5 + x – 3

∴ p(x) = 2x^{4} + 7x^{3} + 14x^{2} + 9x + 2

**Question 10.**

Divide p(x) = x^{3} — 4x^{2} + 5x — 2 by x — 2 Find r(x).

**Answer:**

Here the dividend polynomial p(x) = x^{3} — 4x^{2} + 5x — 2

and the divisor polynomial s(x) = x – 2

Coefficients of x^{3}, x^{2}, x and x^{0} in the dividend polynomial are 1, –4, 5 and –2 respectively.

Equating divisor polynomial x – 2 to 0, we get x = 2.

The reminder obtained when the cubic polynomial x^{3} — 4x^{2} + 5x — 2 is divided by x–2 is r(x) = 0.

**Question 11.**

There are x^{4} + 57x + 15 pens to be distributed in a class of x^{2} + 4x + 2 students. Each student should get the maximum possible number of pens. Find the number of pens received by each student and the number of pens left undistributed (x N).

**Answer:**

Here, dividend polynomial = p(x) = x^{4} + 57x + 15

and divisor polynomial = s(x) = x^{2} + 4x + 2

Thus, the quotient polynomial q(x) = x^{2} – 4x + 14 and the reminder polynomial r(x) = 9x – 13.

⇒ Each student will get (x^{2} – 4x + 14) pens and number of pens left undistributed = 9x – 13

**Question 12.**

A trader bought 2x^{2} — x + 2 TV sets for Rs. 8x^{4} + 7x — 6. Find the price of one TV set.

**Answer:**

Cost of 2x^{2} — x + 2 TV sets = Rs. 8x^{4} + 7x — 6

Here, dividend polynomial = p(x) = 8x^{4} + 7x — 6

and divisor polynomial = s(x) = 2x^{2} — x + 2

⇒ Cost of one TV set = 4x^{2} + 2x – 3.

**Question 13.**

and are two of the zeros of p(x) = 2x^{4} + 7x^{3} — 8x^{2} — 14x + 8. Find the remaining zeros of p(x).

**Answer:**

Given, – √2 and √2 are zeros of p(x).

⇒ (x + √2) and (x – √2) are the factors of p(x).

Thus (x + √2) (x – √2) = (x^{2} – 2) is also a factor of p(x).

To find the remaining zeros, we find the remaining factors using the division process.

Here, dividend polynomial = p(x) = 2x^{4} + 7x^{3} — 8x^{2} — 14x + 8

and divisor polynomial = s(x) = x^{2} – 2

⇒ p(x) = 2x^{4} + 7x^{3} — 8x^{2} — 14x + 8 = (x^{2} – 2)(2x^{2} + 7x — 4)

On factorising 2x^{2} + 7x — 4, we get

2x^{2} + 7x — 4 = 2x^{2} + 8x – x — 4

= 2x (x + 4) –1 (x + 4)

= (2x –1) (x + 4)

Hence the other two zeros of p(x) are and –4.

###### Exercise 2

**Question 1.**State whether the following statements are true or false:

(1) is a zero of the linear polynomial p(x) = 5x + 7.

(2) p(x) = x^{2} + 2x + 1 has two distinct zeros.

(3) The cubic polynomial p(x) = x^{3} + x^{2} — x — 1 has two distinct zeros.

(4) The graph of the cubic polynomial p(x) = x^{3} meets the X—axis at only one point.

(5) Any quadratic polynomial p(x) has at least one zero, x R

**Answer:**(1) p(x) = 5x + 7

∴ p = 5 + 7 = 14

⇒ p ≠0

⇒ is not a zero of p(x).

So the statement is **False**.

(2) Given, p(x) = x^{2} + 2x + 1

To find the zeros of p(x), let p(x) = 0

∴ x^{2} + 2x + 1 = 0

∴ (x + 1)^{2} = 0 Using the identity: (a+ b) ^{2} = (a^{2} + 2ab + b^{2})

∴ x + 1 = 0 or x = –1

∴ x = –1 or x = –1

Here, both the zeros are equal, i.e. –1, and hence not distinct.

So the statement is **False**.

(3) Given, p(x) = x^{3} + x^{2} — x — 1

To find the zeros of p(x), let p(x) = 0

∴ x^{3} + x^{2} — x — 1 = 0

∴ x^{2} (x + 1) – 1 (x + 1) = 0

∴ (x^{2} – 1)(x + 1) = 0

∴ (x – 1) (x + 1)(x + 1) = 0 Using the identity: (a^{2} – b^{2}) = (a – b) (a + b)

∴ x – 1 = 0 or x + 1 = 0 or x + 1 = 0

∴ x = 1 or x = –1 or x = –1

∴ Two distinct zeros of p(x) are 1 and –1.

Hence, p(x) has at the most two distinct zeros.

So the statement is **True**.

(4) Given, p(x) = x^{3}

To find the zeros of p(x), let p(x) = 0

∴x^{3} = 0

⇒ x = 0.

⇒ The graph of p(x) = x^{3} meets the X-axis at only one point i.e. (0, 0).

So the statement is **True**.

(5) If the graph of the quadratic polynomial p(x) does not intersect the x-axis at any point, then the quadratic polynomial does not have any real zero.

So the statement is **False**.

**Question 2.**Find the zeros and number of zeros of p(x) = x^{2} + 9x + 18. Show them on a graph.

**Answer:**Here, p(x) = x^{2} + 9x + 18

To find the zeros of p(x), let p(x) = 0.

∴ x^{2} + 9x + 18 = 0

∴ x^{2} + 6x + 3x + 18 = 0

∴ x(x + 6) + 3(x + 6) = 0

∴ (x + 6)(x + 3) = 0

∴ x = –6 or x = –3

Thus, the number of zeros of p(x) = 2.

To draw the graph of this polynomial, consider the following values of x and corresponding values of p(x):

On plotting the above points on a graph paper we obtain the following shape of the graph:

**Question 3.**Find the zeros, the sum and the product of zeros of p(x) = 4x^{2} + 12x + 5.

**Answer:**Given, p(x) = 4x^{2} + 12x + 5

= 4x^{2} + 10x + 2x + 5

= 2x(2x + 5) + 1(2x + 5)

= (2x + 5)(2x + 1)

To find the zeros of p(x), put p(x) = 0

∴ (2x + 5)(2x + 1) = 0

∴ x = – or x = –

The zeros of p(x) are – and – .

Now, sum of zeros = = = –3

and, product of zeros = – × – =

**Question 4.**—4 and 9 are the sum and product of the zeros respectively of a quadratic polynomial. Find the quadratic polynomial.

**Answer:**Let Î± and Î² be the zeros of the polynomial p(x) = ax^{2} + bx + c.

Given, Î± + Î² = –4 ; Î±Î² = 9

A quadratic polynomial with respect to its roots is given by:

p(x) = x^{2} - (Î± + Î²) x + Î±Î² = 0

Putting the values in this equation we get,

p(x) = x^{2} - (- 4) x + 9

p(x) = x^{2} + 4x + 9

**Question 5.**Find q(x) and r(x), for the quadratic polynomial p(x) = 11x — 21 + 2x^{2} when divided by 1 + 2x

**Answer:**Here, dividend polynomial = p(x) = 11x — 21 + 2x^{2} = 2x^{2} + 11x — 21

and divisor polynomial = s(x) = 1 + 2x = 2x + 1

The quotient polynomial q(x) = x + 5 and the reminder polynomial r(x) = –26.

**Question 6.**Divide 2x^{3} + 3x^{2} — 11x — 6 by x^{2} + x — 6

**Answer:**Here, dividend polynomial = p(x) = 2x^{3} + 3x^{2} — 11x — 6

and divisor polynomial = s(x) = x^{2} + x — 6

Thus, the quotient polynomial q(x) = 2x + 1 and the reminder polynomial r(x) = 0.

**Question 7.**4 is a zero of the cubic polynomial p(x) = x^{3} — 3x^{2} — 6x + 8. Find the remaining zeros of p(x).

**Answer:**Given, 4 is a zero of polynomial p(x).

So (x – 4) is the factor of p(x).

Here, dividend polynomial = p(x) = x^{3} — 3x^{2} — 6x + 8

and divisor polynomial = s(x) = x – 4.

Coefficients of x^{3}, x^{2}, x and x° are 1, –3, –6 and 8 respectively.

Taking x – 4 = 0 we get x = 4

∴ p(x) = x^{3} — 3x^{2} — 6x + 8

= (x – 4) (x^{2} + x – 2)

= (x – 4) (x^{2} – x + 2x – 2)

= (x – 4) (x(x – 1) + 2(x –1))

= (x – 4)(x + 2)(x – 1)

To find the remaining zeros, let p(x) = 0

i.e. (x – 4)(x + 2)(x – 1) = 0

∴ The remaining zeros of p(x) are –2 and 1.

**Question 8.**The product of two polynomials is 3x^{4} + 5x^{3} — 21x^{2} — 53x — 30. If one of them is x^{2} — x — 6, find the other polynomial.

**Answer:**Here p(x) = the dividend polynomial = 3x^{4} + 5x^{3} — 21x^{2} — 53x — 30

s(x) = the divisor polynomial = x^{2} — x — 6

Now,

∴ The quotient polynomial q(x) = 3x^{2} + 8x + 5 and the remainder polynomial r(x) = 0.

∴ The other polynomial is 3x^{2} + 8x + 5.

**Question 9.**2 + √3 and 2 — √3 are the zeros of p(x) = x^{4} — 6x^{3} — 26x^{2} + 138x — 35. Find the remaining zeros of p(x).

**Answer:**Given, 2 + √3 and 2–√3 are zeros of p(x).

⇒ and are the factors of p(x).

Thus = (x^{2} – 4x + 4 –3) = (x^{2} – 4x + 1) is also a factor of p(x).

To find the remaining zeros, we find the remaining factors using the division process.

Here, dividend polynomial = p(x) = 2x^{4} + 7x^{3} — 8x^{2} — 14x + 8

and divisor polynomial = s(x) = x^{2} – 2

⇒ p(x) = 2x^{4} + 7x^{3} — 8x^{2} — 14x + 8 = (x^{2} – 4x + 1)(x^{2} – 2x — 35)

On factorising x^{2} – 2x — 35, we get

x^{2} – 2x — 35 = x^{2} –7x + 5x— 35

= x (x – 7) + 5 (x – 7)

= (x – 7) (x + 5)

Hence the other two zeros of p(x) are 7 and –5.

**Question 10.**The linear polynomial p(x) = 7x — 3 has the zero _____

A.

B.

C. –

D. –

**Answer:**Here, p(x) = 7x — 3

To find the zeros of p(x), consider p(x) = 0

∴ 7x — 3 = 0

∴ x =

Hence, the zero of the given polynomial is .

The correct option is **B**.

**Question 11.**The cubic polynomial p(x) = x^{3} — x has _____ zeros.

A. 0

B. 1

C. 2

D. 3

**Answer:**Here, p(x) = x^{3} — x

To find the zeros of p(x), consider p(x) = 0

∴x^{3} — x = 0

∴ x (x^{2} – 1) = 0

∴ x(x – 1)(x + 1) = 0 Using the identity: (a^{2} – b^{2}) = (a – b) (a + b)

∴ x = 0, x = 1, x = –1

∴ The cubic polynomial has 3 zeros.

The correct option is **D**.

**Question 12.**The graph of p(x) = 3x — 2 — x^{2} intersects the X-axis in _____ points.

A. 0

B. 1

C. 2

D. 3

**Answer:**The zeros of p(x) are the intersections points of the equation, p(x) = 3x – 2 – x^{2} with the x-axis.

∴ The number of distinct zeros of p(x) gives the number of distinct intersection points on the X-axis.

To find the zeros of p(x), consider p(x) = 0

∴3x – 2 – x^{2} = 0

∴ x^{2} – 3x+ 2 = 0

∴ x^{2} – x – 2x+ 2 = 0

∴ x(x – 1) – 2(x – 1) = 0

∴ (x – 1)(x – 2) = 0

∴ x = 1 or x = 2 are the zeros of p(x).

⇒The graph of p(x) intersects X-axis at two points.

The correct option is **C**.

**Question 13.**The sum of the zeros of 3x^{2} + 5x — 2 is _____

A.

B. –

C.

D. –

**Answer:**Given, p(x) = 3x^{2} + 5x — 2

Here a = 3, b = 5, c = –2

Sum of zeros, Î± + Î² = – = –

The correct option is **D**.

**Question 14.**The graph of p(x) = 3x + 5 represents _____

A. a straight line

B. parabola open upwards

C. parabola open downwards

D. a ray

**Answer:**Given, p(x) = 3x+ 5

Since p(x) is a linear polynomial, so its graph is a straight line.

The correct option is **A**.

**Question 15.**A quadratic polynomial has no zero. Its graph _____

A. touches X-axis at any point

B. intersects X-axis at two distinct points

C. does not intersect X-axis at two distinct points

D. is in any one half plane of X-axis

**Answer:**The number of intersection points of a quadratic polynomial at X-axis is the number of its real zeros.

Given, the quadratic polynomial has no zero.

⇒ Its graph does not intersect X-axis or its graph lies in any one half plane of X-axis.

The correct option is **D**.

**Question 16.**For the graph in figure 2.11 y = p(x) has _____ zeros.

A. 1

B. 2

C. 3

D. 4

**Answer:**The graph intersects the X-axis at four distinct points.

⇒ p(x) has four zeros.

The correct option is **D**.

**Question 17.**The product of the zeros of x^{2} — 4x + 3 is _____

A. 1

B. 3

C. 4

D. —4

**Answer:**Given, p(x) = x^{2} — 4x + 3

Here a = 1, b = –4, c = 3

Product of zeros, Î±Î² = = – = 3

The correct option is **B**.

**Question 18.**a = 3, b = 5, c = 7, d = 11 in the standard notation gives the cubic polynomial _____

A. 3x^{3} + 5x^{2} — 7x — 11

B. 3x^{3} — 5x^{2} + 7x — 11

C. 3x^{3} + 5x^{2} — 7x + 11

D. 3x^{3} + 5x^{2} + 7x + 11

**Answer:**The standard form of a cubic polynomial is p(x) = ax^{3} + bx^{2} + cx + d

∴ For a = 3, b = 5, c = 7 and d = 11,

p(x) = 3x^{3} + 5x^{2} + 7x + 11.

The correct option is **D**.

**Question 1.**

State whether the following statements are true or false:

(1) is a zero of the linear polynomial p(x) = 5x + 7.

(2) p(x) = x^{2} + 2x + 1 has two distinct zeros.

(3) The cubic polynomial p(x) = x^{3} + x^{2} — x — 1 has two distinct zeros.

(4) The graph of the cubic polynomial p(x) = x^{3} meets the X—axis at only one point.

(5) Any quadratic polynomial p(x) has at least one zero, x R

**Answer:**

(1) p(x) = 5x + 7

∴ p = 5 + 7 = 14

⇒ p ≠0

⇒ is not a zero of p(x).

So the statement is **False**.

(2) Given, p(x) = x^{2} + 2x + 1

To find the zeros of p(x), let p(x) = 0

∴ x^{2} + 2x + 1 = 0

∴ (x + 1)^{2} = 0 Using the identity: (a+ b) ^{2} = (a^{2} + 2ab + b^{2})

∴ x + 1 = 0 or x = –1

∴ x = –1 or x = –1

Here, both the zeros are equal, i.e. –1, and hence not distinct.

So the statement is **False**.

(3) Given, p(x) = x^{3} + x^{2} — x — 1

To find the zeros of p(x), let p(x) = 0

∴ x^{3} + x^{2} — x — 1 = 0

∴ x^{2} (x + 1) – 1 (x + 1) = 0

∴ (x^{2} – 1)(x + 1) = 0

∴ (x – 1) (x + 1)(x + 1) = 0 Using the identity: (a^{2} – b^{2}) = (a – b) (a + b)

∴ x – 1 = 0 or x + 1 = 0 or x + 1 = 0

∴ x = 1 or x = –1 or x = –1

∴ Two distinct zeros of p(x) are 1 and –1.

Hence, p(x) has at the most two distinct zeros.

So the statement is **True**.

(4) Given, p(x) = x^{3}

To find the zeros of p(x), let p(x) = 0

∴x^{3} = 0

⇒ x = 0.

⇒ The graph of p(x) = x^{3} meets the X-axis at only one point i.e. (0, 0).

So the statement is **True**.

(5) If the graph of the quadratic polynomial p(x) does not intersect the x-axis at any point, then the quadratic polynomial does not have any real zero.

So the statement is **False**.

**Question 2.**

Find the zeros and number of zeros of p(x) = x^{2} + 9x + 18. Show them on a graph.

**Answer:**

Here, p(x) = x^{2} + 9x + 18

To find the zeros of p(x), let p(x) = 0.

∴ x^{2} + 9x + 18 = 0

∴ x^{2} + 6x + 3x + 18 = 0

∴ x(x + 6) + 3(x + 6) = 0

∴ (x + 6)(x + 3) = 0

∴ x = –6 or x = –3

Thus, the number of zeros of p(x) = 2.

To draw the graph of this polynomial, consider the following values of x and corresponding values of p(x):

On plotting the above points on a graph paper we obtain the following shape of the graph:

**Question 3.**

Find the zeros, the sum and the product of zeros of p(x) = 4x^{2} + 12x + 5.

**Answer:**

Given, p(x) = 4x^{2} + 12x + 5

= 4x^{2} + 10x + 2x + 5

= 2x(2x + 5) + 1(2x + 5)

= (2x + 5)(2x + 1)

To find the zeros of p(x), put p(x) = 0

∴ (2x + 5)(2x + 1) = 0

∴ x = – or x = –

The zeros of p(x) are – and – .

Now, sum of zeros = = = –3

and, product of zeros = – × – =

**Question 4.**

—4 and 9 are the sum and product of the zeros respectively of a quadratic polynomial. Find the quadratic polynomial.

**Answer:**

Let Î± and Î² be the zeros of the polynomial p(x) = ax^{2} + bx + c.

Given, Î± + Î² = –4 ; Î±Î² = 9

p(x) = x

^{2}- (Î± + Î²) x + Î±Î² = 0

Putting the values in this equation we get,

p(x) = x^{2} - (- 4) x + 9

p(x) = x^{2} + 4x + 9

**Question 5.**

Find q(x) and r(x), for the quadratic polynomial p(x) = 11x — 21 + 2x^{2} when divided by 1 + 2x

**Answer:**

Here, dividend polynomial = p(x) = 11x — 21 + 2x^{2} = 2x^{2} + 11x — 21

and divisor polynomial = s(x) = 1 + 2x = 2x + 1

The quotient polynomial q(x) = x + 5 and the reminder polynomial r(x) = –26.

**Question 6.**

Divide 2x^{3} + 3x^{2} — 11x — 6 by x^{2} + x — 6

**Answer:**

Here, dividend polynomial = p(x) = 2x^{3} + 3x^{2} — 11x — 6

and divisor polynomial = s(x) = x^{2} + x — 6

Thus, the quotient polynomial q(x) = 2x + 1 and the reminder polynomial r(x) = 0.

**Question 7.**

4 is a zero of the cubic polynomial p(x) = x^{3} — 3x^{2} — 6x + 8. Find the remaining zeros of p(x).

**Answer:**

Given, 4 is a zero of polynomial p(x).

So (x – 4) is the factor of p(x).

Here, dividend polynomial = p(x) = x^{3} — 3x^{2} — 6x + 8

and divisor polynomial = s(x) = x – 4.

Coefficients of x^{3}, x^{2}, x and x° are 1, –3, –6 and 8 respectively.

Taking x – 4 = 0 we get x = 4

∴ p(x) = x^{3} — 3x^{2} — 6x + 8

= (x – 4) (x^{2} + x – 2)

= (x – 4) (x^{2} – x + 2x – 2)

= (x – 4) (x(x – 1) + 2(x –1))

= (x – 4)(x + 2)(x – 1)

To find the remaining zeros, let p(x) = 0

i.e. (x – 4)(x + 2)(x – 1) = 0

∴ The remaining zeros of p(x) are –2 and 1.

**Question 8.**

The product of two polynomials is 3x^{4} + 5x^{3} — 21x^{2} — 53x — 30. If one of them is x^{2} — x — 6, find the other polynomial.

**Answer:**

Here p(x) = the dividend polynomial = 3x^{4} + 5x^{3} — 21x^{2} — 53x — 30

s(x) = the divisor polynomial = x^{2} — x — 6

Now,

∴ The quotient polynomial q(x) = 3x^{2} + 8x + 5 and the remainder polynomial r(x) = 0.

∴ The other polynomial is 3x^{2} + 8x + 5.

**Question 9.**

2 + √3 and 2 — √3 are the zeros of p(x) = x^{4} — 6x^{3} — 26x^{2} + 138x — 35. Find the remaining zeros of p(x).

**Answer:**

Given, 2 + √3 and 2–√3 are zeros of p(x).

⇒ and are the factors of p(x).

Thus = (x^{2} – 4x + 4 –3) = (x^{2} – 4x + 1) is also a factor of p(x).

To find the remaining zeros, we find the remaining factors using the division process.

Here, dividend polynomial = p(x) = 2x^{4} + 7x^{3} — 8x^{2} — 14x + 8

and divisor polynomial = s(x) = x^{2} – 2

⇒ p(x) = 2x^{4} + 7x^{3} — 8x^{2} — 14x + 8 = (x^{2} – 4x + 1)(x^{2} – 2x — 35)

On factorising x^{2} – 2x — 35, we get

x^{2} – 2x — 35 = x^{2} –7x + 5x— 35

= x (x – 7) + 5 (x – 7)

= (x – 7) (x + 5)

Hence the other two zeros of p(x) are 7 and –5.

**Question 10.**

The linear polynomial p(x) = 7x — 3 has the zero _____

A.

B.

C. –

D. –

**Answer:**

Here, p(x) = 7x — 3

To find the zeros of p(x), consider p(x) = 0

∴ 7x — 3 = 0

∴ x =

Hence, the zero of the given polynomial is .

The correct option is **B**.

**Question 11.**

The cubic polynomial p(x) = x^{3} — x has _____ zeros.

A. 0

B. 1

C. 2

D. 3

**Answer:**

Here, p(x) = x^{3} — x

To find the zeros of p(x), consider p(x) = 0

∴x^{3} — x = 0

∴ x (x^{2} – 1) = 0

∴ x(x – 1)(x + 1) = 0 Using the identity: (a^{2} – b^{2}) = (a – b) (a + b)

∴ x = 0, x = 1, x = –1

∴ The cubic polynomial has 3 zeros.

The correct option is **D**.

**Question 12.**

The graph of p(x) = 3x — 2 — x^{2} intersects the X-axis in _____ points.

A. 0

B. 1

C. 2

D. 3

**Answer:**

The zeros of p(x) are the intersections points of the equation, p(x) = 3x – 2 – x^{2} with the x-axis.

∴ The number of distinct zeros of p(x) gives the number of distinct intersection points on the X-axis.

To find the zeros of p(x), consider p(x) = 0

∴3x – 2 – x^{2} = 0

∴ x^{2} – 3x+ 2 = 0

∴ x^{2} – x – 2x+ 2 = 0

∴ x(x – 1) – 2(x – 1) = 0

∴ (x – 1)(x – 2) = 0

∴ x = 1 or x = 2 are the zeros of p(x).

⇒The graph of p(x) intersects X-axis at two points.

The correct option is **C**.

**Question 13.**

The sum of the zeros of 3x^{2} + 5x — 2 is _____

A.

B. –

C.

D. –

**Answer:**

Given, p(x) = 3x^{2} + 5x — 2

Here a = 3, b = 5, c = –2

Sum of zeros, Î± + Î² = – = –

The correct option is **D**.

**Question 14.**

The graph of p(x) = 3x + 5 represents _____

A. a straight line

B. parabola open upwards

C. parabola open downwards

D. a ray

**Answer:**

Given, p(x) = 3x+ 5

Since p(x) is a linear polynomial, so its graph is a straight line.

The correct option is **A**.

**Question 15.**

A quadratic polynomial has no zero. Its graph _____

A. touches X-axis at any point

B. intersects X-axis at two distinct points

C. does not intersect X-axis at two distinct points

D. is in any one half plane of X-axis

**Answer:**

The number of intersection points of a quadratic polynomial at X-axis is the number of its real zeros.

Given, the quadratic polynomial has no zero.

⇒ Its graph does not intersect X-axis or its graph lies in any one half plane of X-axis.

The correct option is **D**.

**Question 16.**

For the graph in figure 2.11 y = p(x) has _____ zeros.

A. 1

B. 2

C. 3

D. 4

**Answer:**

The graph intersects the X-axis at four distinct points.

⇒ p(x) has four zeros.

The correct option is **D**.

**Question 17.**

The product of the zeros of x^{2} — 4x + 3 is _____

A. 1

B. 3

C. 4

D. —4

**Answer:**

Given, p(x) = x^{2} — 4x + 3

Here a = 1, b = –4, c = 3

Product of zeros, Î±Î² = = – = 3

The correct option is **B**.

**Question 18.**

a = 3, b = 5, c = 7, d = 11 in the standard notation gives the cubic polynomial _____

A. 3x^{3} + 5x^{2} — 7x — 11

B. 3x^{3} — 5x^{2} + 7x — 11

C. 3x^{3} + 5x^{2} — 7x + 11

D. 3x^{3} + 5x^{2} + 7x + 11

**Answer:**

The standard form of a cubic polynomial is p(x) = ax^{3} + bx^{2} + cx + d

∴ For a = 3, b = 5, c = 7 and d = 11,

p(x) = 3x^{3} + 5x^{2} + 7x + 11.

The correct option is **D**.