Polynomials Class 10th Mathematics Gujarat Board Solution

Class 10th Mathematics Gujarat Board Solution
Exercise 2.1
  1. Identify the type of the following polynomials: (on base of power) (1) p(x) =…
  2. Obtain the degree of the following polynomials: (1) p(x) = 3x —x^4 + x^2 + 2x^3…
  3. Find the coefficients of the underlined terms: (1) p(x) = 10x^3 + 7 x^2 — 3x + 5…
  4. Obtain the value of the following polynomials at the given values of x: (1) p(x)…
  5. (x + 1) is a factor of p(x) = 3x^3 + 2x^2 + 7 x + 8 Examine the validity of the…
  6. (x + 2) is a factor of p(x) = x^3 + x^2 + x + 2 Examine the validity of the…
  7. (x — 1) is a factor of p(x) = x^4 — 2x^3 + 3x —2 Examine the validity of the…
  8. (x — 3) is a factor of p(x) = x^2 — 2x — 3 Examine the validity of the…
  9. p(x) = x^3 —x^2 —x + 1 Factorize the following polynomials:
  10. p(x) = 5x^2 + 11x + 6 Factorize the following polynomials:
  11. p(x) = x^3 — 3x^2 + 9x — 27 Factorize the following polynomials:
  12. p(x) = x^3 + 2x^2 + 3x + 2 Factorize the following polynomials:
  13. Prove that x — 2 is a factor of p(x) = x^3 — 2x^2
Exercise 2.2
  1. p(x) = x^2 — x Find the number of zeros of the following polynomials:…
  2. p(x) = x —x^2 —1 Find the number of zeros of the following polynomials:…
  3. P(x) = 3x —2 Find the number of zeros of the following polynomials:…
  4. p(x) = x^3 — x Find the number of zeros of the following polynomials:…
  5. Find the number of zeros and real zeros of p(x) = x^3 + 1. Show them by a graph.…
  6. Draw the graph of p(x) = x^2 + 1 and find the real zeros of this polynomial.…
  7. From the figure 2.10 find the number of zeros of y = p(x). (i) (ii) (iii) (iv) (v) (vi)…
  8. Find the number of zeros and zeros of p(x) = x^2 — 4. Represent them…
Exercise 2.3
  1. Prove that 4 and 1 are the zeros of the quadratic polynomial p(x) = x^2 — 5x +…
  2. p(x) = x^2 + 4x — 21 Find the zeros of the following quadratic polynomials:…
  3. p(x) = 6x^2 — 11x + 5 Find the zeros of the following quadratic polynomials:…
  4. p(x) = 4x^2 + 9x + 5 Find the zeros of the following quadratic polynomials:…
  5. p(x) = 3x^2 + 5x — 8 Find the zeros of the following quadratic polynomials:…
  6. p(x) = x^2 — 81 Find the zeros of the following quadratic polynomials:…
  7. p(x) = x^2 — x — 6 Find the zeros of the following quadratic polynomials:…
  8. Find the zeros, the sum of the zeros and the product of the zeros of the…
  9. The sum of zeros = 2; the product of zeros = —3 Obtain a quadratic polynomial…
  10. The sum of zeros = —3; the product of zeros = —4 Obtain a quadratic polynomial…
  11. The sum of zeros = 1/3; the product of zeros = 1/2 Obtain a quadratic…
  12. Obtain the quadratic or the cubic polynomial as the case may be in the standard…
Exercise 2.4
  1. p(x) = 2x^3 — 13x^2 + 23x — 12, s(x) = 2x — 3 Divide the following polynomial…
  2. p(x) = 2/3 x^2 + 5x + 6, s(x) = x + 6 Divide the following polynomial p(x) by…
  3. p(x) = 40x^2 + 11x — 63, s(x) = 8x — 9 Divide the following polynomial p(x) by…
  4. p(x) = 2x^3 + 9x^2 + 13x + 6, s(x) = 2x^2 + 5x + 3 Divide the following…
  5. p(x) = x^4 + 4x^3 + 5x^2 — 7x — 3, s(x) = x^2 — 1 Divide the following…
  6. Find the remainder polynomial when the cubic polynomial x^3 — 3x^2 + 4x + 5 is…
  7. 3 is a zero of p(x) = 3x^3 — x^2 — ax — 45. Find 'a'.
  8. The product of two polynomials is 6x^3 + 29x^2 + 44x + 21 and one of the…
  9. If polynomial p(x) is divided by x^2 + 3x + 5, the quotient polynomial and the…
  10. Divide p(x) = x^3 — 4x^2 + 5x — 2 by x — 2 Find r(x).
  11. There are x^4 + 57x + 15 pens to be distributed in a class of x^2 + 4x + 2…
  12. A trader bought 2x^2 — x + 2 TV sets for Rs. 8x^4 + 7x — 6. Find the price of…
  13. - root 2 and root 2 are two of the zeros of p(x) = 2x^4 + 7x^3 — 8x^2 — 14x + 8.…
Exercise 2
  1. State whether the following statements are true or false: (1) 7/5 is a zero of…
  2. Find the zeros and number of zeros of p(x) = x^2 + 9x + 18. Show them on a graph.…
  3. Find the zeros, the sum and the product of zeros of p(x) = 4x^2 + 12x + 5.…
  4. —4 and 9 are the sum and product of the zeros respectively of a quadratic…
  5. Find q(x) and r(x), for the quadratic polynomial p(x) = 11x — 21 + 2x^2 when…
  6. Divide 2x^3 + 3x^2 — 11x — 6 by x^2 + x — 6
  7. 4 is a zero of the cubic polynomial p(x) = x^3 — 3x^2 — 6x + 8. Find the…
  8. The product of two polynomials is 3x^4 + 5x^3 — 21x^2 — 53x — 30. If one of them…
  9. 2 + √3 and 2 — √3 are the zeros of p(x) = x^4 — 6x^3 — 26x^2 + 138x — 35. Find…
  10. The linear polynomial p(x) = 7x — 3 has the zero _____A. 7/3 B. 3/7 C. - 7/3…
  11. The cubic polynomial p(x) = x^3 — x has _____ zeros.A. 0 B. 1 C. 2 D. 3…
  12. The graph of p(x) = 3x — 2 — x^2 intersects the X-axis in _____ points.A. 0 B.…
  13. The sum of the zeros of 3x^2 + 5x — 2 is _____A. 3/5 B. - 3/5 C. 5/3 D. - 5/3…
  14. The graph of p(x) = 3x + 5 represents _____A. a straight line B. parabola open…
  15. A quadratic polynomial has no zero. Its graph _____A. touches X-axis at any…
  16. For the graph in figure 2.11 y = p(x) has _____ zeros. A. 1 B. 2 C. 3 D. 4…
  17. The product of the zeros of x^2 — 4x + 3 is _____A. 1 B. 3 C. 4 D. —4…
  18. 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) = x2 — 5x + 6

(2) p(x) = x2 — x3 + x + 1

(3) p(x) = 5x2 + 8x + 3

(4) p(x) = x3


Answer:

(1) Here p (x) = x2 — 5x + 6


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


(2) Here p(x) = x2 — x3 + x + 1


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


(3) Here p(x) = 5x2 + 8x + 3


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


(4) Here p(x) = x3


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 —x4 + x2 + 2x3 + 7

(2) p(x) = x3 — 3x —x2 + 6

(3) p(x) = 3x — 9

(4) p(x) = 2x2 — x + 1


Answer:

(1) Here p(x) = 3x —x4 + x2 + 2x3 + 7


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


(2) Here p(x) = x3 — 3x —x2 + 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) = 2x2 — 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) = 10x3+ 7 x2 — 3x + 5

(2) p(x) = 7 — 5x5+ 3x4 + x2 — x

(3) p(x) = 25 — 125x

(4) p(x) = x3 — x2 + x + 7


Answer:

(1) p(x) = 10x3+ 7 x2 — 3x + 5


Coefficient of the underlined term is 10.


(2) p(x) = 7 — 5x5+ 3x4 + x2 — x


Coefficient of the underlined term is –5.


(3) p(x) = 25 — 125x


Coefficient of the underlined term is –125.


(4) p(x) = x3 — x2 + 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) = 2x3 + 3x2 + 7x + 9 ; at x = 0, 1

(2) p(x) = 3x2 + 10x + 7 ; at x = —3, 1

(3) p(x) = x2 — 2x + 5 ; at x = —1, 5

(4) p(x) = 2x4 — 3x3 + 7x + 5 ; at x = —2, 2


Answer:

(1) Given, p(x) = 2x3 + 3x2 + 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) = 3x2 + 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) = x2 — 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) = 2x4 — 3x3 + 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) = 3x3 + 2x2 + 7 x + 8


Answer:

Given, p(x) = 3x3 + 2x2 + 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) = x3 + x2 + x + 2


Answer:

Given, p(x) = x3 + x2 + 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) = x4 — 2x3 + 3x —2


Answer:

Given, p(x) = x4 — 2x3 + 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) = x2 — 2x — 3


Answer:

Given, p(x) = x2 — 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) = x3 —x2—x + 1


Answer:

p(x) = x3 – x2 – x + 1


= x2 (x – 1) – 1(x – 1)


= (x – 1)(x2 – 1)


= (x – 1)(x – 1)(x + 1)


Using the identity: (a2 – b2) = (a – b) (a + b)


= (x – 1)2 (x + 1)



Question 10.

Factorize the following polynomials:

p(x) = 5x2 + 11x + 6


Answer:

p(x) = 5x2 + 11x + 6


= 5x2 + 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) = x3 — 3x2 + 9x — 27


Answer:

p(x) = x3 – 3x2 + 9x – 27


= x2 (x – 3) + 9 (x – 3)


= (x – 3) (x2 + 9)



Question 12.

Factorize the following polynomials:

p(x) = x3 + 2x2 + 3x + 2


Answer:

p(x) = x3 + 2x2 + 3x + 2


= x3 + x2 + x2 + x + 2x + 2 (2x2 = x2 + x2 and 3x = x + 2x)


= x2 (x + 1) + x (x + 1) + 2 (x + 1)


= (x + 1) (x2 + x + 2)



Question 13.

Prove that x — 2 is a factor of p(x) = x3 — 2x2


Answer:

Given, p(x) = x3 — 2x2


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) = x3 — 2x2




Exercise 2.2
Question 1.

Find the number of zeros of the following polynomials:

p(x) = x2 — x


Answer:

Here, p(x) = x2 — x


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


∴ x2 – 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 —x2 —1


Answer:

Here, p(x) = x —x2 —1


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


∴ x – x2 – 1 = 0


Taking negative common


x2 – x + 1 = 0


x2 – x +  + 1 –  = 0


(Adding and subtracting  in order to do factorization)


 +  = 0


Using the identity: (a– b) 2 = (a2 – 2ab + b2)


 = –


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) = x3 — x


Answer:

Here, p(x) = x3 — x


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


∴ x3 – x = 0


∴ x (x2 – 1) = 0


∴ x (x – 1) (x + 1) = 0 Using the identity: (a2 – b2) = (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) = x3 + 1. Show them by a graph.


Answer:

Given, p(x) = x3 + 1


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


∴ x3 + 1 = 0


∴ (x + 1) (x2 – x + 1) = 0 Using the identity (a + b)3 = (a + b) (a2 – ab + b2)


∴ x + 1 = 0 or x2 – x + 1 = 0


∴ x = –1 ; real zeros of x2 – 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) = x3 + 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) = x2 + 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) = x2 — 4. Represent them graphically.


Answer:

Given, p(x) = x2 – 4


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


∴ x2 – 4 = 0


(x – 2)(x + 2) = 0


Using the identity: (a2 – b2) = (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) = x2 — 5x + 4. Also verify the relationship between the zeros and the coefficients of p(x).


Answer:

Here p(x) = x2 – 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) = x2 – 5x + 4.


In p(x) = x2 – 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) = x2 + 4x — 21


Answer:

Here p(x) = x2 + 4x – 21


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


∴ x2 + 4x – 21 = 0


∴ x2 + 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) = 6x2 — 11x + 5


Answer:

Here p(x) = 6x2 — 11x + 5


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


∴6x2 — 11x + 5 = 0


∴ 6x2 – 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) = 4x2 + 9x + 5


Answer:

Here p(x) = 4x2 + 9x + 5


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


∴4x2 + 9x + 5 = 0


∴4x2 + 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) = 3x2 + 5x — 8


Answer:

Here p(x) = 3x2 + 5x — 8


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


∴3x2 + 5x — 8 = 0


∴ 3x2 + 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) = x2 — 81


Answer:

Here p(x) = x2 — 81


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


∴ x2 – 81 = 0


∴ (x)2 – (9)2 = 0


∴ (x – 9) (x + 9) = 0


[Using the identity: (a2 – b2) = (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) = x2 — x — 6


Answer:

Here p(x) = x2 — x — 6


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


∴ x2 — x — 6 = 0


∴ x2 — 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) = 3x2 — x — 4


Answer:

Here p(x) = 3x2 — x — 4


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


∴3x2 — x — 4 = 0


∴ 3x2 + 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 α and β be the zeros of the polynomial p(x) = ax2 + bx + c.


Given, α + β = 2 ; αβ = –3


We know that α + β = – = – . So, –  =  = k, say


Thus, b = –2k, a = k


And αβ = = – . So, c = –3a = –3(k) = –3k


p(x) = ax2 + bx + c


∴ p(x) = (k)x2 + (–2k)x – 3k


∴ p(x) = k (x2 – 2x – 3)



Question 10.

Obtain a quadratic polynomial with the following conditions:

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


Answer:

Let α and β be the zeros of the polynomial p(x) = ax2 + bx + c.


Given, α + β = –3 ; αβ = –4


We know that α + β = – = – . So, –  =  = k, say


Thus, b = –3k, a = k


And αβ = = – . So, c = –4a = –4(k) = –4k


p(x) = ax2 + bx + c


∴ p(x) = (k)x2 + (–3k)x – 4k


∴ p(x) = k (x2 – 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 α and β be the zeros of the polynomial p(x) = ax2 + bx + c.


Given, α + β =  ; αβ = 


We know that α + β = – = . So, –  =  = k, say


Thus, b = –k, a = 3k


And αβ = = . So, c =  = 


p(x) = ax2 + bx + c


∴ p(x) = (3k)x2 + (–k)x + 


∴ p(x) = k (3x2 – x + )


∴ p(x) =  (6x2 – 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) = ax2 + bx + c; a ≠ 0, a, b, c ε R.


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


p(x) = 6x2 + 17x + 11.


(2) The cubic polynomial is,


p(x) = ax3 + bx2 + cx + d; a ≠ 0, a, b, c, d ε R.


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


p(x) = x3 – x2 – x + 1.


(3) The quadratic polynomial is,


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


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


p(x) = 5x2 + 7x + 2.


(4) The cubic polynomial is,


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


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


p(x) = x3 – 3x2 – x + 3.


(5) The cubic polynomial is,


p(x) = ax3 + bx2 + cx + d; a ≠ 0, a, b, c, d ε R.


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


p(x) = 3x3 – 5x2 – 11x – 3.




Exercise 2.4
Question 1.

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

p(x) = 2x3 — 13x2 + 23x — 12, s(x) = 2x — 3


Answer:

Here, dividend polynomial = p(x) = 2x3 – 13x2 + 23x – 12


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



Thus, the quotient polynomial q(x) = x2 – 5x + 4 and the reminder polynomial r(x) = 0.



Question 2.

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

p(x) = x2 + 5x + 6, s(x) = x + 6


Answer:

Here, dividend polynomial = p(x) = x2 + 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) = 40x2 + 11x — 63, s(x) = 8x — 9


Answer:

Here, dividend polynomial = p(x) = 40x2 + 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) = 2x3 + 9x2 + 13x + 6, s(x) = 2x2 + 5x + 3


Answer:

Here, dividend polynomial = p(x) = 2x3 + 9x2 + 13x + 6


and divisor polynomial = s(x) = 2x2 + 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) = x4 + 4x3 + 5x2 — 7x — 3, s(x) = x2 — 1


Answer:

Here, dividend polynomial = p(x) = 2x3 – 13x2 + 23x – 12


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



Thus, the quotient polynomial q(x) = x2 + 4x + 6 and the reminder polynomial r(x) = –3x + 3.



Question 6.

Find the remainder polynomial when the cubic polynomial x3 — 3x2 + 4x + 5 is divided by x — 2


Answer:

Here the dividend polynomial p(x) = x3 — 3x2 + 4x + 5 and the divisor polynomial s(x) = x – 2


Coefficients of x3, x2, x and x0 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 x3 — 3x2 + 4x + 5 is divided by x–2 is r(x) = 9.



Question 7.

3 is a zero of p(x) = 3x3 — x2 — 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 6x3 + 29x2 + 44x + 21 and one of the polynomials is 3x + 7. Find the other polynomial.


Answer:

Here p(x) = the dividend polynomial = 6x3 + 29x2 + 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 2x2 + 5x + 3



Question 9.

If polynomial p(x) is divided by x2 + 3x + 5, the quotient polynomial and the remainder polynomials are 2x2 + x + 1 and x — 3 respectively. Find p(x).


Answer:

Here, divisor polynomial s(x) = x2 + 3x + 5


quotient polynomial q(x) = 2x2 + 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) = (x2 + 3x + 5) × (2x2 + x + 1) + (x – 3)


∴ p(x) = 2x4 + x3 + x2 + 6x3 + 3x2 + 3x + 10x2 + 5x + 5 + x – 3


∴ p(x) = 2x4 + 7x3 + 14x2 + 9x + 2



Question 10.

Divide p(x) = x3 — 4x2 + 5x — 2 by x — 2 Find r(x).


Answer:

Here the dividend polynomial p(x) = x3 — 4x2 + 5x — 2


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


Coefficients of x3, x2, x and x0 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 x3 — 4x2 + 5x — 2 is divided by x–2 is r(x) = 0.



Question 11.

There are x4 + 57x + 15 pens to be distributed in a class of x2 + 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) = x4 + 57x + 15


and divisor polynomial = s(x) = x2 + 4x + 2



Thus, the quotient polynomial q(x) = x2 – 4x + 14 and the reminder polynomial r(x) = 9x – 13.


⇒ Each student will get (x2 – 4x + 14) pens and number of pens left undistributed = 9x – 13



Question 12.

A trader bought 2x2 — x + 2 TV sets for Rs. 8x4 + 7x — 6. Find the price of one TV set.


Answer:

Cost of 2x2 — x + 2 TV sets = Rs. 8x4 + 7x — 6


Here, dividend polynomial = p(x) = 8x4 + 7x — 6


and divisor polynomial = s(x) = 2x2 — x + 2



⇒ Cost of one TV set = 4x2 + 2x – 3.



Question 13.

 and are two of the zeros of p(x) = 2x4 + 7x3 — 8x2 — 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) = (x2 – 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) = 2x4 + 7x3 — 8x2 — 14x + 8


and divisor polynomial = s(x) = x2 – 2



⇒ p(x) = 2x4 + 7x3 — 8x2 — 14x + 8 = (x2 – 2)(2x2 + 7x — 4)


On factorising 2x2 + 7x — 4, we get


2x2 + 7x — 4 = 2x2 + 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) = x2 + 2x + 1 has two distinct zeros.

(3) The cubic polynomial p(x) = x3 + x2 — x — 1 has two distinct zeros.

(4) The graph of the cubic polynomial p(x) = x3 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) = x2 + 2x + 1


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


∴ x2 + 2x + 1 = 0


∴ (x + 1)2 = 0 Using the identity: (a+ b) 2 = (a2 + 2ab + b2)


∴ 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) = x3 + x2 — x — 1


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


∴ x3 + x2 — x — 1 = 0


∴ x2 (x + 1) – 1 (x + 1) = 0


∴ (x2 – 1)(x + 1) = 0


∴ (x – 1) (x + 1)(x + 1) = 0 Using the identity: (a2 – b2) = (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) = x3


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


∴x3 = 0


⇒ x = 0.


⇒ The graph of p(x) = x3 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) = x2 + 9x + 18. Show them on a graph.


Answer:

Here, p(x) = x2 + 9x + 18


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


∴ x2 + 9x + 18 = 0


∴ x2 + 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) = 4x2 + 12x + 5.


Answer:

Given, p(x) = 4x2 + 12x + 5


= 4x2 + 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 α and β be the zeros of the polynomial p(x) = ax2 + bx + c.

Given, α + β = –4 ; αβ = 9

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

p(x) = x2 - (α + β) x + αβ = 0

Putting the values in this equation we get,

p(x) = x2 - (- 4) x + 9

p(x) = x2 + 4x + 9


Question 5.

Find q(x) and r(x), for the quadratic polynomial p(x) = 11x — 21 + 2x2 when divided by 1 + 2x


Answer:

Here, dividend polynomial = p(x) = 11x — 21 + 2x2 = 2x2 + 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 2x3 + 3x2 — 11x — 6 by x2 + x — 6


Answer:

Here, dividend polynomial = p(x) = 2x3 + 3x2 — 11x — 6


and divisor polynomial = s(x) = x2 + 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) = x3 — 3x2 — 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) = x3 — 3x2 — 6x + 8


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


Coefficients of x3, x2, x and x° are 1, –3, –6 and 8 respectively.


Taking x – 4 = 0 we get x = 4



∴ p(x) = x3 — 3x2 — 6x + 8


= (x – 4) (x2 + x – 2)


= (x – 4) (x2 – 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 3x4 + 5x3 — 21x2 — 53x — 30. If one of them is x2 — x — 6, find the other polynomial.


Answer:

Here p(x) = the dividend polynomial = 3x4 + 5x3 — 21x2 — 53x — 30


s(x) = the divisor polynomial = x2 — x — 6


Now,



∴ The quotient polynomial q(x) = 3x2 + 8x + 5 and the remainder polynomial r(x) = 0.


∴ The other polynomial is 3x2 + 8x + 5.



Question 9.

2 + √3 and 2 — √3 are the zeros of p(x) = x4 — 6x3 — 26x2 + 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  = (x2 – 4x + 4 –3) = (x2 – 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) = 2x4 + 7x3 — 8x2 — 14x + 8


and divisor polynomial = s(x) = x2 – 2



⇒ p(x) = 2x4 + 7x3 — 8x2 — 14x + 8 = (x2 – 4x + 1)(x2 – 2x — 35)


On factorising x2 – 2x — 35, we get


x2 – 2x — 35 = x2 –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) = x3 — x has _____ zeros.
A. 0

B. 1

C. 2

D. 3


Answer:

Here, p(x) = x3 — x


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


∴x3 — x = 0


∴ x (x2 – 1) = 0


∴ x(x – 1)(x + 1) = 0 Using the identity: (a2 – b2) = (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 — x2 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 – x2 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 – x2 = 0


∴ x2 – 3x+ 2 = 0


∴ x2 – 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 3x2 + 5x — 2 is _____
A. 

B. –

C. 

D. –


Answer:

Given, p(x) = 3x2 + 5x — 2


Here a = 3, b = 5, c = –2


Sum of zeros, α + β = –  = – 


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 x2 — 4x + 3 is _____
A. 1

B. 3

C. 4

D. —4


Answer:

Given, p(x) = x2 — 4x + 3


Here a = 1, b = –4, c = 3


Product of zeros, αβ =  = –  = 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. 3x3 + 5x2 — 7x — 11

B. 3x3 — 5x2 + 7x — 11

C. 3x3 + 5x2 — 7x + 11

D. 3x3 + 5x2 + 7x + 11


Answer:

The standard form of a cubic polynomial is p(x) = ax3 + bx2 + cx + d


∴ For a = 3, b = 5, c = 7 and d = 11,


p(x) = 3x3 + 5x2 + 7x + 11.


The correct option is D.