# Ncert class 10 polynomial solutions

Are you searching for class 10 chapter – 2 polynomial solutions? you’re at the right place. You can get accurate CBSE NCERT Solutions for all subjects. These solutions will help you a lot in scoring good marks in the exams.

Page No: 28

**Polynomial solutions – Exercises 2.1 **

1. The graphs of y = p(x) are given in the following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

**Answer**

(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.

[the_ad id=”619″]

Page No: 33

**Polynomial solutions – Exercise 2.2**

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

(i) *x*^{2} – 2*x* – 8

(ii) 4*s*^{2} – 4*s* + 1

(iii) 6*x*^{2} – 3 – 7*x*

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

*t*

^{2}– 15

*x*

^{2}–

*x*– 4

**Answer**

(i) *x*^{2} – 2*x* – 8

= (*x* – 4) (*x* + 2)

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

Therefore, the zeroes of *x*^{2} – 2*x* – 8 are 4 and -2.

*x*)/Coefficient of

*x*

^{2}

*x*

^{2}

(ii) 4*s*^{2} – 4*s* + 1

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

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

^{2}– 4s + 1 are 1/2 and 1/2.

Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient of *s)*/Coefficient of *s*^{2}

Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of *s*^{2}.

(iii) 6*x*^{2} – 3 – 7*x*

*= *6*x*^{2 }– 7*x *– 3

= (3*x* + 1) (2*x* – 3)

The value of 6*x*^{2} – 3 – 7*x* is zero when 3*x* + 1 = 0 or 2*x* – 3 = 0, i.e., *x* = -1/3 or *x* = 3/2

*x*

^{2}– 3 – 7

*x*are -1/3 and 3/2.

Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of *x*)/Coefficient of *x*^{2}

Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of *x*^{2}.

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

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

= 4*u*(*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 = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of *u*)/Coefficient of *u*^{2}

Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of *u*^{2}.

(v) *t*^{2} – 15

= *t*^{2 }– 0.*t* – 15

= (*t *– √15) (*t* + √15)

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 = √15 + -√15 = 0 = -0/1 = -(Coefficient of *t*)/Coefficient of *t*^{2}

Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of *u*^{2}.

(vi) 3*x*^{2} –* x* – 4

= (3*x* – 4) (*x* + 1)

The value of 3*x*^{2} –* x* – 4 is zero when 3*x* – 4 = 0 and *x* + 1 = 0,i.e., when *x* = 4/3 or *x* = -1

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

*x*)/Coefficient of

*x*

^{2}

Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of *x*^{2}.

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

(i) 1/4 , -1

**Answer**

(i) 1/4 , -1

Let the polynomial be *ax*^{2} + *bx* + *c*, and its zeroes be α and ß

α + ß = 1/4 = –*b*/*a*

αß = -1 = -4/4 = *c*/*a*

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

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

(ii) √2 , 1/3

Let the polynomial be *ax*^{2} + *bx* + *c*, and its zeroes be α and ß

α + ß = √2 = 3√2/3 = –*b*/*a*

αß = 1/3 = *c*/*a*

If *a* = 3, then *b* = -3√2, *c* = 1

Therefore, the quadratic polynomial is 3*x*^{2} -3√2*x* +1.

(iii) 0, √5

Let the polynomial be *ax*^{2} + *bx* + *c*, and its zeroes be α and ß

α + ß = 0 = 0/1 = –*b*/*a*

αß = √5 = √5/1 = *c*/*a*

If *a* = 1, then *b* = 0, *c* = √5

Therefore, the quadratic polynomial is *x*^{2} + √5.

(iv) 1, 1

Let the polynomial be *ax*^{2} + *bx* + *c*, and its zeroes be α and ß

α + ß = 1 = 1/1 = –*b*/*a*

αß = 1 = 1/1 = *c*/*a*

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

Therefore, the quadratic polynomial is *x*^{2} – *x* +1.

(v) -1/4 ,1/4

Let the polynomial be *ax*^{2} + *bx* + *c*, and its zeroes be α and ß

α + ß = -1/4 = –*b*/*a*

αß = 1/4 = *c*/*a*

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

Therefore, the quadratic polynomial is 4*x*^{2} + *x* +1.

(vi) 4,1

Let the polynomial be *ax*^{2} + *bx* + *c*, and its zeroes be α and ß

α + ß = 4 = 4/1 = –*b*/*a*

αß = 1 = 1/1 = *c*/*a*

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

Therefore, the quadratic polynomial is *x*^{2} – 4*x* +1.

Page No: 36

[the_ad id=”619″]

Download NCERT chapter 2 – Polynomial solution –Download Now

**Polynomial solutions – Exercise 2.3**

1. Divide the polynomial *p*(*x*) by the polynomial *g*(*x*) and find the quotient and remainder in each of the following:

^{ }

**Answer**

(i) *p*(*x*) = *x*^{3} – 3*x*^{2} + 5*x* – 3, *g*(*x*) = *x*^{2} – 2

Quotient = *x*-3 and remainder 7*x* – 9

*p*(

*x*) =

*x*

^{4}– 3

*x*

^{2}+ 4x + 5,

*g*(

*x*) =

*x*

^{2}+ 1 –

*x*

Quotient = *x*^{2} + *x *– 3 and remainder 8

*p*(

*x*) =

*x*

^{4}– 5

*x*+ 6,

*g*(

*x*) = 2 –

*x*

^{2}

Quotient = –*x*^{2} -2 and remainder -5*x* +10

^{ [the_ad id=”619″]}

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

second polynomial by the first polynomial:

**Answer**

(i) *t*^{2} – 3, 2*t*^{4} + 3*t*^{3} – 2*t*^{2} – 9*t* – 12

*t*^{2} – 3 exactly divides 2*t*^{4} + 3*t*^{3} – 2*t*^{2} – 9*t* – 12 leaving no remainder. Hence, it is a factor of 2*t*^{4} + 3*t*^{3} – 2*t*^{2} – 9*t* – 12.

(ii) *x*^{2} + 3*x* + 1, 3*x*^{4} + 5*x*^{3} – 7*x*^{2} + 2*x* + 2

*x*

^{2}+ 3

*x*+ 1 exactly divides 3

*x*

^{4}+ 5

*x*

^{3}– 7

*x*

^{2}+ 2

*x*+ 2 leaving no remainder. Hence, it is factor of 3

*x*

^{4}+ 5

*x*

^{3}– 7

*x*

^{2}+ 2

*x*+ 2.

(iii) *x*^{3} – 3*x* + 1, *x*^{5} – 4*x*^{3} + *x*^{2} + 3*x* + 1

*x*

^{3}– 3

*x*+ 1 didn’t divides exactly

*x*

^{5}– 4

*x*

^{3}+

*x*

^{2}+ 3

*x*+ 1 and leaves 2 as remainder. Hence, it not a factor of

*x*

^{5}– 4

*x*

^{3}+

*x*

^{2}+ 3

*x*+ 1.

*x*

^{4}+ 6

*x*

^{3}– 2

*x*

^{2}– 10

*x*– 5, if two of its zeroes are √(5/3)

and – √(5/3).

**Answer**

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

Since the two zeroes are √(5/3) and – √(5/3).

As it has the term (*x *+ 1)^{2} , therefore, there will be 2 zeroes at *x* = – 1.

Hence, the zeroes of the given polynomial are √(5/3) and – √(5/3), – 1 and – 1.

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

-2*x* + 4, respectively. Find *g*(*x*).

**Answer**

Here in the given question,

*x*

^{3}– 3

*x*

^{2}+

*x*+ 2

*x*– 2

*x*+ 4

*g*(

*x*)

*x*

^{3}– 3

*x*

^{2}+

*x*+ 2 = (

*x*– 2) ×

*g*(

*x*) + (-2

*x*+ 4)⇒

*x*

^{3}– 3

*x*

^{2}+

*x*+ 2 – (-2

*x*+ 4) = (

*x*– 2) ×

*g*(

*x*)

⇒

*x*

^{3}– 3

*x*

^{2}+ 3

*x*– 2 = (

*x*– 2) ×

*g*(

*x*)

⇒

*g*(

*x*) = (

*x*

^{3}– 3

*x*

^{2}+ 3

*x*– 2)/(

*x*– 2)

∴ *g*(*x*) = (*x*^{2} – *x* + 1)

5.Give examples of polynomial *p*(*x*), *g*(*x*), *q*(*x*) and *r*(*x*), which satisfy the division algorithm and

(i) deg *p*(*x*) = deg *q*(*x*)

(ii) deg *q*(*x*) = deg *r*(*x*)

(iii) deg *r*(*x*) = 0

**Answer**

*x*

^{2}+ 2

*x*+ 2 by 2

Here,

*p*(

*x*) = 6

*x*

^{2}+ 2

*x*+ 2

*g*(

*x*) = 2

*q*(

*x*) = 3

*x*

^{2}+

*x*+ 1

*r*(

*x*) = 0

Degree of

*p*(

*x*) and

*q*(

*x*) is same i.e. 2.

Checking for division algorithm,

*p*(

*x*) =

*g*(

*x*) ×

*q*(

*x*) +

*r*(

*x*)

Or, 6

*x*

^{2}+ 2

*x*+ 2 = 2

*x*(3

*x*

^{2}+

*x*+ 1)

Hence, 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., 1.

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

Page No: 36

Read also – Real number solution

Download NCERT chapter 2 – Polynomial solution –Download Now

**Polynomial solutions – Exercise 2.4 (Optional)**

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) 2*x*^{3} + *x*^{2} – 5*x *+ 2; 1/2, 1, -2

(ii) *x*^{3} – 4*x*^{2} + 5*x –* 2; 2, 1, 1

**Answer**

(i) *p*(*x*) = 2*x*^{3} + *x*^{2} – 5*x *+ 2

Now for zeroes, putting the given value in x.

*p*(*1/2*) = 2(1/2)^{3} + *(1/2)*^{2} – 5(1/2)+ 2

= (2×1/8) + 1/4 – 5/2 + 2

= 1/4 + 1/4 – 5/2 + 2

= 1/2 – 5/2 + 2 = 0

*p*(*1*) = 2(1)^{3} + *(1)*^{2} – 5(1)+ 2

= (2×1) + 1 – 5 + 2

= 2 + 1 – 5 + 2 = 0

*p*(*-2*) = 2(-2)^{3} + (-2)^{2} – 5(-2)+ 2

= (2 × -8) + 4 + 10 + 2

= -16 + 16 = 0

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

Comparing the given polynomial with a*x*^{3} + b*x*^{2} + c*x *+ d, we get a=2, b=1, c=-5, d=2

Also, α=1/2, β=1 and γ=-2

Now,

-b/a = α+β+γ

⇒ 1/2 = 1/2 + 1 – 2

⇒ 1/2 = 1/2

c/a = αβ+βγ+γα

⇒ -5/2 = (1/2 × 1) + (1 × -2) + (-2 × 1/2)

⇒ -5/2 = 1/2 – 2 – 1

⇒ -5/2 = -5/2

-d/a = αβγ

⇒ -2/2 = (1/2 × 1 × -2)

⇒ -1 = 1

Thus, the relationship between zeroes and the coefficients are verified.

[the_ad id=”619″]

(ii) *p*(*x*) = *x*^{3} – 4*x*^{2} + 5*x –* 2

Now for zeroes, putting the given value in x.

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

= 8 – 16 + 10 – 2

= 0

*p*(*1*) = 1^{3} – 4*(1)*^{2} + 5(1)- 2

= 1 – 4 + 5 – 2

= 0

*p*(*1*) = 1^{3} – 4*(1)*^{2} + 5(1)- 2

= 1 – 4 + 5 – 2

= 0

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

Comparing the given polynomial with a*x*^{3} + b*x*^{2} + c*x *+ d, we get a=1, b=-4, c=5, d=-2

Also, α=2, β=1 and γ=1

Now,

-b/a = α+β+γ

⇒ 4/1 = 2 + 1 + 1

⇒ 4 = 4

c/a = αβ+βγ+γα

⇒ 5/1 = (2 × 1) + (1 × 1) + (1 × 2)

⇒ 5 = 2 + 1 + 2

⇒ 5 = 5

-d/a = αβγ

⇒ 2/1 = (2 × 1 × 1)

⇒ 2 = 2

Thus, the relationship between zeroes and the coefficients are verified.

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**

Let the polynomial be *ax*^{3} + *bx*^{2 }+* cx + d *and the zeroes be α, β and γ

Then, α + β + γ = -(-2)/1 = 2 = -b/a

αβ + βγ + γα = -7 = -7/1 = c/a

αβγ = -14 = -14/1 = -d/a

∴ a = 1, b = -2, c = -7 and d = 14

So, one cubic polynomial which satisfy the given conditions will be *x*^{3} – 2*x*^{2 }–* 7x + 14*

3. If the zeroes of the polynomial *x*^{3} – 3*x*^{2} + *x* + 1 are a–b, a, a+b, find a and b.

**Answer**

Since, (a – b), a, (a + b) are the zeroes of the polynomial *x*^{3} – 3*x*^{2} + *x* + 1.

Therefore, sum of the zeroes = (a – b) + a + (a + b) = -(-3)/1 = 3

⇒ 3a = 3 ⇒ a =1

∴ Sum of the products of is zeroes taken two at a time = a(a – b) + a(a + b) + (a + b) (a – b) =1/1 = 1

a^{2} – ab + a^{2} + ab + a^{2} – b^{2 }= 1

⇒ 3a^{2} – b^{2} =1

Putting the value of a,

⇒ 3(1)^{2} – b^{2} = 1

⇒ 3 – b^{2} = 1

⇒ b^{2} = 2

⇒ b = ±√2

Hence, a = 1 and b = ±√2

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

**Answer**

2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x^{4} – 6x^{3} – 26x^{2} + 138x – 35.

Let x = 2±√3

So, x-2 = ±√3

On squaring, we get x^{2} – 4x + 4 = 3,

⇒ x^{2} – 4x + 1= 0

Now, dividing p(x) by x^{2} – 4x + 1

∴ p(x) = x^{4} – 6x^{3} – 26x^{2} + 138x – 35

= (x^{2} – 4x + 1) (x^{2} – 2x – 35)

= (x^{2} – 4x + 1) (x^{2} – 7x + 5x – 35)

= (x^{2} – 4x + 1) [x(x – 7) + 5 (x – 7)]

= (x^{2} – 4x + 1) (x + 5) (x – 7)

∴ (x + 5) and (x – 7) are other factors of p(x).

∴ – 5 and 7 are other zeroes of the given polynomial.

^{4}– 6x

^{3}+ 16x

^{2}– 25x + 10 is divided by another polynomial x

^{2}– 2x + k, the remainder comes out to be x + a, find k and a.

**Answer**

^{4}– 6x

^{3}+ 16x

^{2}– 25x + 10 by x

^{2}– 2x + k

## Download NCERT chapter 2 – Polynomial solution –Download Now

**Go Back To NCERT Solutions for Class 10th Maths**

**Disclaimer – These questions are taken from Ncert official website.**