##### Class 10^{th} Mathematics CBSE Solution

**Exercise 2.1****Exercise 2.2**- Find the zeroes of the following quadratic polynomials and verify the…
- Find a quadratic polynomial each with the given numbers as the sum and product…

**Exercise 2.3**- Divide the polynomial by the polynomial and find the quotient and remainder in…
- Check whether the first polynomial is a factor of the second polynomial by…
- Obtain all other zeroes of 3x4+ 6x3- 2x2- 10x - 5,if two of its zeroes areand…
- On dividing x^3 - 3x^2 + x + 2 by a polynomial g(x) the quotient and remainder…
- Give examples of polynomials and which satisfy the division algorithm and (i)…

**Exercise 2.4**- Verify that the numbers given alongside of the cubic polynomials below are their…
- Find a cubic polynomial with the sum, sum of the product of its zeroes taken two…
- If the zeroes of the polynomial are (a - b) and (a + b). Find a and b.…
- If two zeroes of the polynomial x^4 - 6x^3 - 26x^2 + 138x - 35are find other…
- If the polynomial x^4 - 6x^3 + 16x^2 - 25x+10 is divided by another polynomial…

**Exercise 2.1**

**Exercise 2.2**

- Find the zeroes of the following quadratic polynomials and verify the…
- Find a quadratic polynomial each with the given numbers as the sum and product…

**Exercise 2.3**

- Divide the polynomial by the polynomial and find the quotient and remainder in…
- Check whether the first polynomial is a factor of the second polynomial by…
- Obtain all other zeroes of 3x4+ 6x3- 2x2- 10x - 5,if two of its zeroes areand…
- On dividing x^3 - 3x^2 + x + 2 by a polynomial g(x) the quotient and remainder…
- Give examples of polynomials and which satisfy the division algorithm and (i)…

**Exercise 2.4**

- Verify that the numbers given alongside of the cubic polynomials below are their…
- Find a cubic polynomial with the sum, sum of the product of its zeroes taken two…
- If the zeroes of the polynomial are (a - b) and (a + b). Find a and b.…
- If two zeroes of the polynomial x^4 - 6x^3 - 26x^2 + 138x - 35are find other…
- If the polynomial x^4 - 6x^3 + 16x^2 - 25x+10 is divided by another polynomial…

###### Exercise 2.1

**Question 1.**The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials, p(x). Find the number of zeroes of p(x) in each case.

**Answer:**The number of zeroes for any graph is the number of values of x for which y is equal to zero. And y is equal to zero at the point where a graph cuts x axis.

So, to find the number of zeroes of a polynomial, watch the number of times it cuts the x axis.

(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.

(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.

(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.

(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.

(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

**Question 1.**

The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials, p(x). Find the number of zeroes of p(x) in each case.

**Answer:**

The number of zeroes for any graph is the number of values of x for which y is equal to zero. And y is equal to zero at the point where a graph cuts x axis.

So, to find the number of zeroes of a polynomial, watch the number of times it cuts the x axis.

(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.

(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.

(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.

(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.

(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

###### Exercise 2.2

**Question 1.**Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x^{2 }- 2x - 8

(ii) 4s^{2 }- 4s + 1

(iii) 6x^{2 }- 3 - 7x

(iv) 4u^{2 }+ 8u

(v) t^{2 }- 15

(vi) 3x^{2 }- x - 4

**Answer:**Zeroes of the polynomial are the values of the variable of the polynomial when the polynomial is put equal to zero.

Let p(x) be a polynomial with any number of terms any number of degree.

Now, zeroes of the polynomial will be the values of x at which p(x) = 0.

If p(x) = ax^{2 }+ bx + c is a quadratic polynomial (highest power is equal to 2) and its roots are Î± and Î², then

Sum of the roots = Î± + Î² = - b/a

Product of roots = Î±Î² = c/a

**(i)** p(x) = x^{2 }- 2x - 8

So, the zeroes will be the values of x at which p(x) = 0

Therefore,

⇒ x^{2} - 4x + 2x - 8 = 0

(We will factorize 2 such that the product of the factors is equal to 8 and difference is equal to 2)

⇒ x(x - 4) + 2(x - 4) = 0

= (x - 4)(x + 2)

The value of x^{2 }- 2x - 8 is zero when *x* − 4 = 0 or *x* + 2 = 0,

i.e, *x* = 4 or *x* = −2

Therefore, The zeroes of x^{2 }- 2x - 8 are 4 and −2.

Sum of zeroes = 4 + (-2) = 2

Hence, it is verified that,

Product of zeroes =

Hence, it is verified that,

**(ii)** 4s^{2 }- 4s + 1

= (2s)^{2} - 2(2s)1 + 1^{2}

As, we know (a - b)^{2} = a^{2} - 2ab + b^{2}, the above equation can be written as

= (2s - 1)^{2}

The value of 4*s*^{2} − 4*s* + 1 is zero when 2*s* − 1 = 0, when, s = 1/2 , 1/2.

Therefore, the zeroes of 4*s*^{2} − 4*s* + 1 are and .

Sum of zeroes =

Product of zeroes =

Hence Verified.

**(iii)**6x^{2 }- 3 - 7x

= 6x^{2} - 7x - 3

(We will factorize 7 such that the product of the factors is equal to 18 and the difference is equal to - 7)

= 6x^{2} + 2x - 9x - 3

= 2x(3x + 1) - 3(3x + 1)

= (3x + 1)(2x - 3)

The value of 6*x*^{2} − 3 − 7*x* is zero when 3*x* + 1 = 0 or 2*x* − 3 = 0,

i.e.

Therefore, the zeroes of 6*x*^{2} − 3 − 7*x* are .

Sum of zeroes =

Product of zeroes =

Hence, verified.

**(iv)** 4u^{2 }+ 8u

= 4u^{2 }+ 8u + 0

= 4u (u + 2)

The value of 4*u*^{2} + 8*u* is zero when 4*u* = 0 or *u* + 2 = 0,

i.e., *u* = 0 or *u* = −2

Therefore, the zeroes of 4*u*^{2} + 8*u* are 0 and −2.

Sum of zeroes =

Product of zeroes =

**(v)** t^{2} – 15

= t^{2} – (√15)^{2}

= (t - √15)(t + √15) [As, x^{2 }- y^{2} = (x - y)(x + y)]

The value of *t*^{2} − 15 is zero when (t - √15) = 0 or (t + √15) = 0,

i.e., when t = √15 or t = -√15

Therefore, the zeroes of *t*^{2} − 15 are √15 and -√15.

Sum of zeroes =

Product of zeroes =

Hence verified.

**(vi)** 3x^{2} – x – 4

(We will factorize 1 in such a way that the product of factors is equal to 12 and the difference is equal to 1)

= 3x^{2} - 4x + 3x - 4

= x(3x - 4) + 1(3x - 4)

= (3x – 4 )(x + 1)

The value of 3*x*^{2} − *x* − 4 is zero when 3*x* − 4 = 0 or *x* + 1 = 0,

when

Therefore, the zeroes of 3*x*^{2} − *x* − 4 are and -1

Sum of zeroes =

Product of zeroes =

Hence, verified.

**Question 2.**Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

**Answer:**

(i) , -1

we know that for a quadratic equation in the form ax^{2} + bx + c = 0, and its zerors are Î± and Î², then

sum of zeroes is

and product of zeroes is

Let the polynomial be , then

Let a = 4, then b = -1, c = -4

Therefore, the quadratic polynomial is 4*x*^{2} − *x* − 4.

(ii)

we know that for a quadratic equation in the form ax^{2} + bx + c = 0, and its zerors are Î± and Î², then

sum of zeroes is

and product of zeroes is

Let the polynomial be , then

and

If a = 3, then b = , and c = 1

Therefore, the quadratic polynomial is

(iii) 0,

we know that for a quadratic equation in the form ax^{2} + bx + c = 0, and its zerors are Î± and Î², then

sum of zeroes is

and product of zeroes is

Let the polynomial be , then

If a = 1, then b = 0, c =

Therefore, the quadratic polynomial is .

(iv) 1, 1

we know that for a quadratic equation in the form ax^{2} + bx + c = 0, and its zerors are Î± and Î², then

sum of zeroes is

and product of zeroes is

Let the polynomial be , then

If a = 1, then b = -1, c = 1

Therefore, the quadratic polynomial is .

(v)

we know that for a quadratic equation in the form ax^{2} + bx + c = 0, and its zerors are Î± and Î², then

sum of zeroes is

and product of zeroes is

Let the polynomial be , then

If a = 4, then b = 1, c = 1

Therefore, the quadratic polynomial is .

(vi) 4, 1

we know that for a quadratic equation in the form ax^{2} + bx + c = 0, and its zerors are Î± and Î², then

sum of zeroes is

and product of zeroes is

Let the polynomial be , then

If a = 1, then b = -4, c = 1

Therefore, the quadratic polynomial is .

**Question 1.**

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x^{2 }- 2x - 8

(ii) 4s^{2 }- 4s + 1

(iii) 6x^{2 }- 3 - 7x

(iv) 4u^{2 }+ 8u

(v) t^{2 }- 15

(vi) 3x^{2 }- x - 4

**Answer:**

Zeroes of the polynomial are the values of the variable of the polynomial when the polynomial is put equal to zero.

Let p(x) be a polynomial with any number of terms any number of degree.

Now, zeroes of the polynomial will be the values of x at which p(x) = 0.

If p(x) = ax^{2 }+ bx + c is a quadratic polynomial (highest power is equal to 2) and its roots are Î± and Î², then

Sum of the roots = Î± + Î² = - b/a

Product of roots = Î±Î² = c/a**(i)** p(x) = x^{2 }- 2x - 8

So, the zeroes will be the values of x at which p(x) = 0

Therefore,

⇒ x^{2} - 4x + 2x - 8 = 0

(We will factorize 2 such that the product of the factors is equal to 8 and difference is equal to 2)

⇒ x(x - 4) + 2(x - 4) = 0

= (x - 4)(x + 2)

The value of x^{2 }- 2x - 8 is zero when *x* − 4 = 0 or *x* + 2 = 0,

i.e, *x* = 4 or *x* = −2

Therefore, The zeroes of x^{2 }- 2x - 8 are 4 and −2.

Sum of zeroes = 4 + (-2) = 2

Hence, it is verified that,

Product of zeroes =

Hence, it is verified that,

**(ii)** 4s^{2 }- 4s + 1

= (2s)^{2} - 2(2s)1 + 1^{2}

As, we know (a - b)^{2} = a^{2} - 2ab + b^{2}, the above equation can be written as

= (2s - 1)^{2}

The value of 4*s*^{2} − 4*s* + 1 is zero when 2*s* − 1 = 0, when, s = 1/2 , 1/2.

Therefore, the zeroes of 4*s*^{2} − 4*s* + 1 are and .

Sum of zeroes =

Product of zeroes =

Hence Verified.

**(iii)**6x^{2 }- 3 - 7x

= 6x^{2} - 7x - 3

(We will factorize 7 such that the product of the factors is equal to 18 and the difference is equal to - 7)

= 6x^{2} + 2x - 9x - 3

= 2x(3x + 1) - 3(3x + 1)

= (3x + 1)(2x - 3)

The value of 6*x*^{2} − 3 − 7*x* is zero when 3*x* + 1 = 0 or 2*x* − 3 = 0,

i.e.

Therefore, the zeroes of 6*x*^{2} − 3 − 7*x* are .

Sum of zeroes =

Product of zeroes =

Hence, verified.

**(iv)** 4u^{2 }+ 8u

= 4u^{2 }+ 8u + 0

= 4u (u + 2)

The value of 4*u*^{2} + 8*u* is zero when 4*u* = 0 or *u* + 2 = 0,

i.e., *u* = 0 or *u* = −2

Therefore, the zeroes of 4*u*^{2} + 8*u* are 0 and −2.

Sum of zeroes =

Product of zeroes =

**(v)** t^{2} – 15

= t^{2} – (√15)^{2}

= (t - √15)(t + √15) [As, x^{2 }- y^{2} = (x - y)(x + y)]

The value of *t*^{2} − 15 is zero when (t - √15) = 0 or (t + √15) = 0,

i.e., when t = √15 or t = -√15

Therefore, the zeroes of *t*^{2} − 15 are √15 and -√15.

Sum of zeroes =

Product of zeroes =

Hence verified.

**(vi)** 3x^{2} – x – 4

(We will factorize 1 in such a way that the product of factors is equal to 12 and the difference is equal to 1)

= 3x^{2} - 4x + 3x - 4

= x(3x - 4) + 1(3x - 4)

= (3x – 4 )(x + 1)

The value of 3*x*^{2} − *x* − 4 is zero when 3*x* − 4 = 0 or *x* + 1 = 0,

when

Therefore, the zeroes of 3*x*^{2} − *x* − 4 are and -1

Sum of zeroes =

Product of zeroes =

Hence, verified.

**Question 2.**

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

**Answer:**

(i) , -1

^{2}+ bx + c = 0, and its zerors are Î± and Î², then

sum of zeroes is

and product of zeroes is

Let the polynomial be , then

Let a = 4, then b = -1, c = -4

Therefore, the quadratic polynomial is 4*x*^{2} − *x* − 4.

(ii)

we know that for a quadratic equation in the form ax^{2}+ bx + c = 0, and its zerors are Î± and Î², then

sum of zeroes is

and product of zeroes is

Let the polynomial be , then

If a = 3, then b = , and c = 1

Therefore, the quadratic polynomial is

(iii) 0,

we know that for a quadratic equation in the form ax^{2}+ bx + c = 0, and its zerors are Î± and Î², then

sum of zeroes is

and product of zeroes is

Let the polynomial be , then

If a = 1, then b = 0, c =

Therefore, the quadratic polynomial is .

(iv) 1, 1

we know that for a quadratic equation in the form ax^{2}+ bx + c = 0, and its zerors are Î± and Î², then

sum of zeroes is

and product of zeroes is

Let the polynomial be , then

If a = 1, then b = -1, c = 1

Therefore, the quadratic polynomial is .

(v)

we know that for a quadratic equation in the form ax^{2}+ bx + c = 0, and its zerors are Î± and Î², then

sum of zeroes is

and product of zeroes is

Let the polynomial be , then

If a = 4, then b = 1, c = 1

Therefore, the quadratic polynomial is .

(vi) 4, 1

we know that for a quadratic equation in the form ax^{2}+ bx + c = 0, and its zerors are Î± and Î², then

sum of zeroes is

and product of zeroes is

Let the polynomial be , then

If a = 1, then b = -4, c = 1

Therefore, the quadratic polynomial is .

###### Exercise 2.3

**Question 1.**Divide the polynomial by the polynomial and find the quotient and remainder in each of the following :

(i)

(ii)

(iii)

**Answer:**

(i)By long division method we have,

Quotient = *x* − 3

Remainder = 7*x* − 9

(ii) By long division method we have,

Quotient = *x*^{2} + *x* − 3

Remainder = 8

(iii) By long division method we have,

Quotient = −*x*^{2} − 2

Remainder = −5*x* +10

**Question 2.**Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i)

(ii)

(iii)

**Answer:**

(i) t^{2}-3 = t^{2}+0t-3

Since the remainder is 0,

Hence, t^{2} – 3 is a factor of 2t^{4}+3t^{3}-2t^{2}-9t-12.

_{(ii)}

Since the remainder is 0,

Hence, x^{2}+3x+1 is a factor of 3x^{4}+5x^{3}-7x^{2}+2x+2.

_{(iii)}

Since the remainder ≠0,

Hence, x^{3}-3x+1 is not a factor of x^{5}-4x^{3}+ x^{2}+3x+1.

**Question 3.**Obtain all other zeroes of 3x^{4 }+ 6x^{3 }- 2x^{2 }- 10x - 5, if two of its zeroes areand

**Answer:**p(x) = 3x^{4 }+ 6x^{3 }- 2x^{2 }- 10x - 5

Since the two zeroes are .

is a factor of 3x^{4 }+ 6x^{3 }- 2x^{2 }- 10x - 5.

Therefore, we divide the given polynomial by .

We know,

Dividend = (Divisor × quotient) + remainder

3 x^{4 }+ 6 x^{3 }- 2x^{2 }- 10 x - 5 =

3 x^{4 }+ 6 x^{3 }- 2 x^{2 }- 10 x - 5 =

As (a+b)^{2} = a^{2} + b^{2} + 2ab

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

3 x^{4 }+ 6 x^{3 }- 2 x^{2 }- 10 x - 5 = 3 () (x + 1)^{2}

Therefore, its zero is given by *x* + 1 = 0.

*⇒ x* = −1,-1

**Hence, the zeroes of the given polynomial are and - 1 , -1.**

**Question 4.**On dividing x^{3} - 3x^{2} + x + 2 by a polynomial g(x) the quotient and remainder were (x - 2) and (-2x + 4), respectively. Find g(x).

**Answer:**Given,

Polynomial, p(x) = x^{3} - 3x^{2} + x + 2 (dividend)

Quotient = (*x* − 2)

Remainder = (− 2*x* + 4)

To find : divisor = g(x)

we know,

**Dividend = Divisor × Quotient + Remainder**

⇒ x^{3} - 3x^{2} + x +2 = g(x) × (x - 2) + (-2x + 4)

⇒ x^{3} - 3x^{2} + x + 2 + 2x - 4 = g(x)(x - 2)

⇒ x^{3} - 3x^{2} + 3x - 2 = g(x)(x - 2)

*g*(*x*) is the quotient when we divide (x^{3} - 3x^{2} + 3x - 2) by (x - 2)

**Question 5.**Give examples of polynomials and which satisfy the division algorithm and

(i)

(ii)

(iii)

**Answer:**Degree of a polynomial is the highest power of the variable in the polynomial. For example if f(x) = x^{3} - 2x^{2 }+ 1, then the degree of this polynomial will be 3.

**(i)** By division Algorithm : p(x) = g(x) x q(x) + r(x)

It means when P(x) is divided by g(x) then quotient is q(x) and remainder is r(x)

We need to start with p(x) = q(x)

This means that the degree of polynomial p(x) and quotient q(x) is same. This can only happen if the degree of g(x) = 0 i.e p(x) is divided by a constant

Let p(x) = x^{2} + 1 and g(x) = 2

The,

Clearly, Degree of p(x) = Degree of q(x)

2. Checking for division algorithm,

*p*(*x*) = *g*(*x*) × *q*(*x*) + *r*(*x)*

=

Thus, the division algorithm is satisfied.

**(ii)** Let us assume the division of *x*^{3}*+ x* by *x*^{2},

Here*,*

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

*g*(*x*) = *x*^{2}

*q*(*x*) = *x* and *r*(*x*) = *x*

Clearly, the degree of *q*(*x*) and *r*(*x*) is the same i.e.,

Checking for division algorithm,

*p*(*x*) = *g*(*x*) × *q*(*x*) + *r*(*x) x*^{3}*+ x*

= (*x*^{2} ) × *x* + *x x*^{3}*+ x = x*^{3}*+ x*

Thus, the division algorithm is satisfied.

**(iii)** Degree of the remainder will be 0 when the remainder comes to a constant.

Let us assume the division of *x*^{3}*+* 1by *x*^{2}.

Here*,*

*p*(*x*) = *x*^{3}*+* 1 *g*(*x*) = *x*^{2}

*q*(*x*) = *x* and *r*(*x*) = 1

Clearly, the degree of *r*(*x*) is 0. Checking for division algorithm,

*p*(*x*) = *g*(*x*) × *q*(*x*) + *r*(*x)x*^{3}*+* 1

= (*x*^{2} ) × *x* + 1 *x*^{3}*+* 1 *= x*^{3}*+*1

Thus, the division algorithm is satisfied.

**Question 1.**

Divide the polynomial by the polynomial and find the quotient and remainder in each of the following :

(i)

(ii)

(iii)

**Answer:**

(i)By long division method we have,

Quotient = *x* − 3

Remainder = 7*x* − 9

(ii) By long division method we have,

Quotient = *x*^{2} + *x* − 3

Remainder = 8

(iii) By long division method we have,

Quotient = −*x*^{2} − 2

Remainder = −5*x* +10

**Question 2.**

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i)

(ii)

(iii)

**Answer:**

(i) t^{2}-3 = t^{2}+0t-3

Since the remainder is 0,

Hence, t^{2} – 3 is a factor of 2t^{4}+3t^{3}-2t^{2}-9t-12.

_{(ii)}

Since the remainder is 0,

Hence, x^{2}+3x+1 is a factor of 3x^{4}+5x^{3}-7x^{2}+2x+2.

_{(iii)}

Since the remainder ≠0,

Hence, x^{3}-3x+1 is not a factor of x^{5}-4x^{3}+ x^{2}+3x+1.

**Question 3.**

Obtain all other zeroes of 3x^{4 }+ 6x^{3 }- 2x^{2 }- 10x - 5, if two of its zeroes areand

**Answer:**

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

Since the two zeroes are .

is a factor of 3x^{4 }+ 6x^{3 }- 2x^{2 }- 10x - 5.

Therefore, we divide the given polynomial by .

We know,

Dividend = (Divisor × quotient) + remainder

3 x^{4 }+ 6 x^{3 }- 2x^{2 }- 10 x - 5 =

3 x^{4 }+ 6 x^{3 }- 2 x^{2 }- 10 x - 5 =

As (a+b)^{2} = a^{2} + b^{2} + 2ab

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

^{2}

3 x^{4 }+ 6 x^{3 }- 2 x^{2 }- 10 x - 5 = 3 () (x + 1)^{2}

Therefore, its zero is given by *x* + 1 = 0.*⇒ x* = −1,-1

**Hence, the zeroes of the given polynomial are and - 1 , -1.**

**Question 4.**

On dividing x^{3} - 3x^{2} + x + 2 by a polynomial g(x) the quotient and remainder were (x - 2) and (-2x + 4), respectively. Find g(x).

**Answer:**

Given,

Polynomial, p(x) = x^{3} - 3x^{2} + x + 2 (dividend)

Quotient = (*x* − 2)

Remainder = (− 2*x* + 4)

To find : divisor = g(x)

we know,

**Dividend = Divisor × Quotient + Remainder**

⇒ x

^{3}- 3x

^{2}+ x +2 = g(x) × (x - 2) + (-2x + 4)

⇒ x^{3} - 3x^{2} + x + 2 + 2x - 4 = g(x)(x - 2)

⇒ x^{3} - 3x^{2} + 3x - 2 = g(x)(x - 2)

*g*(*x*) is the quotient when we divide (x^{3} - 3x^{2} + 3x - 2) by (x - 2)

**Question 5.**

Give examples of polynomials and which satisfy the division algorithm and

(i)

(ii)

(iii)

**Answer:**

Degree of a polynomial is the highest power of the variable in the polynomial. For example if f(x) = x^{3} - 2x^{2 }+ 1, then the degree of this polynomial will be 3.

**(i)** By division Algorithm : p(x) = g(x) x q(x) + r(x)

It means when P(x) is divided by g(x) then quotient is q(x) and remainder is r(x)

We need to start with p(x) = q(x)

This means that the degree of polynomial p(x) and quotient q(x) is same. This can only happen if the degree of g(x) = 0 i.e p(x) is divided by a constant

Let p(x) = x^{2} + 1 and g(x) = 2

The,

Clearly, Degree of p(x) = Degree of q(x)

2. Checking for division algorithm,

*p*(*x*) = *g*(*x*) × *q*(*x*) + *r*(*x)*

=

Thus, the division algorithm is satisfied.

**(ii)** Let us assume the division of *x*^{3}*+ x* by *x*^{2},

Here*,*

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

*g*(*x*) = *x*^{2}

*q*(*x*) = *x* and *r*(*x*) = *x*

Clearly, the degree of *q*(*x*) and *r*(*x*) is the same i.e.,

Checking for division algorithm,

*p*(*x*) = *g*(*x*) × *q*(*x*) + *r*(*x) x*^{3}*+ x*

= (*x*^{2} ) × *x* + *x x*^{3}*+ x = x*^{3}*+ x*

Thus, the division algorithm is satisfied.

**(iii)** Degree of the remainder will be 0 when the remainder comes to a constant.

Let us assume the division of *x*^{3}*+* 1by *x*^{2}.

Here*,*

*p*(*x*) = *x*^{3}*+* 1 *g*(*x*) = *x*^{2}

*q*(*x*) = *x* and *r*(*x*) = 1

Clearly, the degree of *r*(*x*) is 0. Checking for division algorithm,

*p*(*x*) = *g*(*x*) × *q*(*x*) + *r*(*x)x*^{3}*+* 1

= (*x*^{2} ) × *x* + 1 *x*^{3}*+* 1 *= x*^{3}*+*1

Thus, the division algorithm is satisfied.

###### Exercise 2.4

**Question 1.**Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i)

(ii)

**Answer:**(i) P(x) =

Now for zeroes, putting the given values in x.

P(1/2) = 2(1/2)^{3} + (1/2)^{2} - 5(1/2) + 2

= (1/4) + (1/4) - (5/2) + 2

= (1 + 1 - 10 + 8)/2

= 0/2 = 0

P(1) =

P(-2) =

Thus, 1/2, 1 and -2 are zeroes of given polynomial.

Comparing given polynomial with ax^{3} + bx^{2} + cx + d and Taking zeroes as Î±, Î², and Î³, we have

Now, We know the relation between zeroes and the coefficient of a standard cubic polynomial as

Substituting value, we have

Since, LHS = RHS (Relation Verified)

Since LHS = RHS, Relation verified.

Since LHS = RHS, Relation verified.

Thus, all three relationships between zeroes and the coefficient is verified.

(ii) p(x) = x^{3} – 4x^{2} + 5x – 2

Now for zeroes , put the given value in x.

P(2) = =

P(1) =

P(1) =

Thus, 2, 1 , 1 are the zeroes of the given polynomial.

Now,

Comparing the given polynomial with ax^{3} + bx^{2} + cx + d, we get

Now,

4 = 4

5 = 5

Î±Î²Î³ =

2 × 1 × 1 = 2

2 = 2

Thus, all three relationships between zeroes and the coefficient is verified.

**Question 2.**Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.

**Answer:**For a cubic polynomial equation, ax^{3} + bx^{2} + cx + d, and zeroes Î±, Î² and Î³

we know that

Let the polynomial be ax^{3} + bx^{2} + cx + d, and zeroes Î±, Î² and Î³.

A cubic polynomial with respect to its zeroes is given by,

x^{3} - (sum of zeroes) x^{2} + (Sum of the product of roots taken two at a time) x - Product of Roots = 0

x^{3} - (2) x^{2} + (- 7) x - (- 14) = 0

x^{3} - (2) x^{2} + (- 7) x + 14 = 0

Hence, the polynomial is x^{3} - 2x^{2} - 7x + 14.

**Question 3.**If the zeroes of the polynomial are (a - b), a and (a + b). Find a and b.

**Answer:**Given,

p(x) =

zeroes are = a – b , a + b , a

C

=

Sum of zeroes =

a =

The zeroes are = (1 - b), 1 and (1 + b)

Product of zeroes = (1 - b)(1 + b)

(1 - b)(1 + b) = -q/m

So,

We get,

**Question 4.**If two zeroes of the polynomial x^{4} - 6x^{3} - 26x^{2} + 138x - 35 are find other zeroes.

**Answer:**Given:

2+√3 and 2-√3 are zeroes of given equation,

Therefore,

(x - 2 + √3)(x - 2 - √3) should be a factor of given equation.

Also, (x - 2 + √3)(x - 2 - √3) = x^{2} - 2x - √3x -2x + 4 + 2√3 + √3x - 2√3 - 3

= x^{2} - 4x + 1

To find other zeroes, we divide given equation by x^{2} - 4x + 1

We get ,

x^{4} - 6x^{3} - 26x^{2} + 138x - 35 = (x^{2} - 4x + 1)(x^{2} - 2x - 35)

Now factorizing x^{2} - 2 x - 35 we get,

The value of polynomial is also zero when ,

x - 7 = 0

or

x = 7

And, x + 5 = 0

or

x = -5

**Hence, 7 and -5 are also zeroes of this polynomial.**

**Question 5.**If the polynomial x^{4} - 6x^{3} + 16x^{2} - 25x + 10 is divided by another polynomial x^{2} - 2k + k the remainder comes out to be x + a, find k and a.

**Answer:****To solve this question divide x**^{4} - 6 x^{3} + 16 x^{2} - 25 x + 10 by x^{2} - 2 x + k by long division method

Let us divide, by

So, remainder = (2k - 9)x + (10 - 8k + k^{2})

But given remainder = x + a

⇒ (2k - 9)x + (10 - 8k + k^{2}) = x + a

Comparing coefficient of x, we have

2k - 9 = 1

⇒ 2k = 10

⇒ k = 5

and

Comparing constant term,

10 - 8k + k^{2} = a

⇒ a = 10 - 8(5) + 5^{2}

⇒ a = 10 - 40 + 25

⇒ a = -5

So, the value of k is 5 and a is -5.

**Question 1.**

Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i)

(ii)

**Answer:**

(i) P(x) =

Now for zeroes, putting the given values in x.

P(1/2) = 2(1/2)^{3} + (1/2)^{2} - 5(1/2) + 2

= (1/4) + (1/4) - (5/2) + 2

= (1 + 1 - 10 + 8)/2

= 0/2 = 0

P(1) =

P(-2) =

Thus, 1/2, 1 and -2 are zeroes of given polynomial.

Comparing given polynomial with ax^{3} + bx^{2} + cx + d and Taking zeroes as Î±, Î², and Î³, we have

Now, We know the relation between zeroes and the coefficient of a standard cubic polynomial as

Substituting value, we have

Since LHS = RHS, Relation verified.

Thus, all three relationships between zeroes and the coefficient is verified.

(ii) p(x) = x^{3} – 4x^{2} + 5x – 2

Now for zeroes , put the given value in x.

P(2) = =

P(1) =

P(1) =

Thus, 2, 1 , 1 are the zeroes of the given polynomial.

Now,

Comparing the given polynomial with ax^{3} + bx^{2} + cx + d, we get

4 = 4

5 = 5

Î±Î²Î³ =

2 × 1 × 1 = 2

2 = 2

Thus, all three relationships between zeroes and the coefficient is verified.

**Question 2.**

Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.

**Answer:**

For a cubic polynomial equation, ax^{3} + bx^{2} + cx + d, and zeroes Î±, Î² and Î³

we know that

Let the polynomial be ax^{3} + bx^{2} + cx + d, and zeroes Î±, Î² and Î³.

A cubic polynomial with respect to its zeroes is given by,

x^{3} - (sum of zeroes) x^{2} + (Sum of the product of roots taken two at a time) x - Product of Roots = 0

x^{3} - (2) x^{2} + (- 7) x - (- 14) = 0

x^{3} - (2) x^{2} + (- 7) x + 14 = 0

Hence, the polynomial is x^{3} - 2x^{2} - 7x + 14.

**Question 3.**

If the zeroes of the polynomial are (a - b), a and (a + b). Find a and b.

**Answer:**

Given,

p(x) =

zeroes are = a – b , a + b , a

C

=

Sum of zeroes =

a =

The zeroes are = (1 - b), 1 and (1 + b)

Product of zeroes = (1 - b)(1 + b)

(1 - b)(1 + b) = -q/m

So,

We get,

**Question 4.**

If two zeroes of the polynomial x^{4} - 6x^{3} - 26x^{2} + 138x - 35 are find other zeroes.

**Answer:**

Given:

2+√3 and 2-√3 are zeroes of given equation,

Therefore,

(x - 2 + √3)(x - 2 - √3) should be a factor of given equation.

Also, (x - 2 + √3)(x - 2 - √3) = x^{2} - 2x - √3x -2x + 4 + 2√3 + √3x - 2√3 - 3

= x^{2} - 4x + 1

^{2}- 4x + 1

We get ,

x^{4} - 6x^{3} - 26x^{2} + 138x - 35 = (x^{2} - 4x + 1)(x^{2} - 2x - 35)

Now factorizing x

^{2}- 2 x - 35 we get,

The value of polynomial is also zero when ,

x - 7 = 0or

x = 7

And, x + 5 = 0

or

x = -5

**Hence, 7 and -5 are also zeroes of this polynomial.**

**Question 5.**

If the polynomial x^{4} - 6x^{3} + 16x^{2} - 25x + 10 is divided by another polynomial x^{2} - 2k + k the remainder comes out to be x + a, find k and a.

**Answer:**

**To solve this question divide x ^{4} - 6 x^{3} + 16 x^{2} - 25 x + 10 by x^{2} - 2 x + k by long division method**

Let us divide, by

So, remainder = (2k - 9)x + (10 - 8k + k^{2})

But given remainder = x + a

⇒ (2k - 9)x + (10 - 8k + k^{2}) = x + a

Comparing coefficient of x, we have

2k - 9 = 1

⇒ 2k = 10

⇒ k = 5

and

Comparing constant term,

10 - 8k + k^{2} = a

⇒ a = 10 - 8(5) + 5^{2}

⇒ a = 10 - 40 + 25

⇒ a = -5

So, the value of k is 5 and a is -5.