**NCERT Solutions Class ****10** **Maths**** Chapter ****2** **Polynomials**

**NCERT Solutions Class ****10** **Maths**** Chapter ****2** **Polynomials****:-** **National Council of Educational Research and Training Class 10 Maths Chapter 2 Solutions – Polynomials. NCERT Solutions Class 10 Maths Chapter 2 PDF Download.**

**NCERT Solutions Class ****10** **Maths**** Chapter ****2****: Overview**

Board |
NCERT |

Class |
10 |

Subject |
Maths |

Chapter |
2 |

Chapter Name |
Polynomials |

Topic |
Exercise Solutions |

**NCERT Solutions **

**Class ****10** **Maths**

**Chapter ****2**** – ****Polynomials**

__Exercise – 2.1__

__Exercise – 2.1__

**(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.**

**Solution:**

i)

The number of zeroes is O. As the graph does not intersects at any point of x axis.

(ii)

The number of zeroes is 1. As the graph intersects the x-axis at one point only.

(iii)

The number of zeroes is 3 as the graph interests the x-axis at 3 points.

(iv)

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

(v)

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

(vi)

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

** **

__Exercise 2.2__

__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} – 2x – 8**

**Solution: **

x^{2} – 2x – 8 = 0

Or, x^{2} – (4 – 2) x – 8 = 0

Or, x^{2} – 4x + 2x – 8 = 0

Or, x (x – 4) + 2 (x – 4) = 0

Or, (x – 4 (x + 2) = 0

x – 4 = 0

Or, x = 4

x + 2 = 0

Or, x = -2

The zeroes of x^{2} – 2x – 8 are 4 and -2.

Now, Sum of zeroes = 4 + (-2) = 2/1 = – (Coefficient of)/(coefficient of)

= – (-2)/1 = 2

Product of zeroes = (4) × (-2) = -8

= constant term/coefficient of x^{2}

= -8/1 =-8

** **

**(ii) 4s ^{2} – 4s + 1**

**Solution: **

4s^{2} – 4S + 1 = 0

Or, 4s^{2} – (2 + 2) S + 1 = 0

Or, 4s^{2} – 2s – 2s + 1 = 0

Or, 2s (2s – 1) -1 (2s – 1) = 0

Or, (2s – 1) (2s – 1) = 0

2s – 1 = 0

S = 1/2

2s – 1 = 0

s = 1/2

So, the zeroes of 4s^{2} – 4s + 1 are 1/2 and 1/2.

Now, sum of zeroes = 1/2 + 1/2 = 1 = ½

= – (coefficient of s)/Coefficient of s^{2}

= – (-4)/4 = 1

Product of zeroes = ½ × 1/2 = 1/4 = constant terms/coefficient of s^{2}

= 1/4

** **

**(iii) 6x ^{2} – 3 – 7x**

**Solution: **

6x^{2} – 3 – 7x = 0

Or, 6x^{2} – 7x – 3 = 0

Or, 6x^{2} – (9 – 2) x – 3 = 0

Or, 6x^{2} – 9x + 2x – 3 = 0

Or, 3x (2x – 3) + 1 (2x – 3) = 0

Or, (3x + 1) (2x – 3) = 0

2x – 3 = 0

2x = 3

x = 3/2

3x + 1 = 0

3x = -1

x = – 1/3

The zeroes of 6x^{2} – 3 – 7x are (3/2) and (-1/3)

Sum of zeroes = 3/2 + (-1/3) = 9-2/6 = 7/6

= (coefficient of x)/coefficient of x^{2}

= – (-7)/6 = 7/6

Product of zeroes = (3/2) × (-1/3)

= – 1/2 constant term/coefficient of x^{2} = -3/6 = – 1/2

** **

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

**Solution: **

4u^{2} + 8u = 0

Or, u^{2} + 2u = 0

Or, u (u + 2) = 0

u = 0

u + 2 = 0

u = -2

The zeroes of 4u^{2} + 8u are 0 and –2

Now, Sum of zeroes = 0 + (-2) = -2/1 = – (Coefficient of u)/(Coefficient of u^{2})

= -8/4 = -2

Product of zeroes = 0×(-2 = 0

= constant term/coefficient of u^{2}

= 0/8 = 0

** **

**(v) t ^{2} – 15 **

**Solution: **

t^{2} – 15 = 0

Or, t^{2} = 15

Or, t = ± √15

Or, t = + √15

t = – √15

The zeroes of t^{2} – 15 are √15 and (-√15)

Now, Sum of zeroes = + √15 – √15 = 0 – (coefficient t)/(Coefficient of t^{2})

= 0/1 = 0

Product of zeroes = (√15) × (-√15)

= -15 = constant term/coefficient of t^{2} = -15/1 = -15

** **

**(vi) 3x ^{2} – x – 4 **

**Solution: **

3x^{2} – x – 4 = 0

Or, 3x^{2} – (4 – 3) x – 4 = 0

Or, 3x^{2} – 4x + 3x – 4 = 0

Or, x (3x – 4) + 1 (3x – 4) = 0

Or, (x + 1) (3x – 4) = 0

x + 1 = 0

x = -1

3x – 4 = 0

3x = 4

x = 4/3

The zeroes of (3x^{2} – x – 4) are (-1) and 4/3

Now, Sum of zeroes = -1 + 4/3 = -3+4/3

= 1/3 = – (coefficient of x)/coefficient of x^{2}

= – (-1)/3 = 1/3

Product of zeroes = (-1) × 4/3

= – 4/3 constant term/coefficient of x^{2}

= -4/3

** **

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

**(i) 1/4, -1 **

**Solution: **

Let, the quadratic polynomial be x^{2} – (α + β) x + αβ = 0 and its zeroes be α and β

We have, α + β = 1/4 and α.β = -1

Now, x^{2} – (α + β) x + α.β = 0

Or, x^{2} – 1/4 x + (-1) = 0

Or, 4x^{2} – x – 4 = 0

∴ The quadratic polynomial be 4x^{2} – x – 4

** **

**(ii) ****√2, 1/3**

**Solution: **

Let, the quadratic polynomial be x^{2} – (α+β) x + αβ = 0 and it’s zeroes be α and β.

We have α + β = √2 and αβ = 1/3

Now, x^{2} – (α + β) x + α.β = 0

Or, x^{2} – √2 x + 1/3 = 0

Or, 3x^{2} – 3√2 x + 1 = 0

∴ The quadratic polynomial be 3x^{2} – 3√2 x + 1

** **

**(iii) 0, √5 **

**Solution: **

Let, the quadratic polynomial be x^{2 }– (α + β) x + α.β = 0 and it’s zeroes be α and β.

We have α + β = 0 and α.β = √5

Now, x^{2} – (α + β) x + α.β = 0

Or, x^{2} – 0.x + √5 = 0

Or, x^{2} + √5 = 0

The quadratic polynomial be x^{2} + √5 = 0

** **

**(iv) 1, 1 **

**Solution: **

Let, the quadratic polynomial be

x^{2} – (α + β) x + α.β = 0, and it’s zeroes be α and β.

We have (α + β) = 1 and α.β = 1

Now, x^{2} – (α + β) x + α.β = 0

x^{2} – 1.x – 1 = 0

x^{2} – x + 1 = 0

∴ The quadratic polynomial be x^{2} – x + 1

** **

**(****v) – ¼, 1/4 **

**Solution: **

Let, the quadratic polynomial be x^{2} – (α + β) x + α.β = 0. and it zeroes be α and β

We have α + β = – 1/4 and α.β = 1/4

Now, x^{2} – (α + β) x + α.β = 0

x^{2} + x/4 + ¼ = 0

4x^{2} + x + 1 = 0

∴ The quadratic polynomial is 4x^{2} + x + 1

** **

**(vi) 4, 1 **

**Solution: **

Let, the quadratic polynomial be x^{2} – (α + β) x + α.β = 0 and it zeroes be α and β.

We have α + β = 4 and α.β = 1

Now, x^{2 }– (α + β) x + α.β = 0

Or, x^{2} – 4x + 1 = 0

The quadratic polynomial be x^{2} – 4x + 1.

** **

__EXERCISE 2.3__

__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:**

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

**Solution: **

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

Hence, Quotient = x – 3, Remainder = 7x – 9

** **

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

**Solution: **

Given, p(x) = x^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} + 1 – x

Hence, we get Quotient = x^{2} + x – 3

Remainder = 8.

** **

**(iii) p(x) = x ^{4} – 5x + 6, g(x) = 2 – x^{2}**

**Solution: **

p(x) = x^{4} – 5x + 6, g(x) = 2-x^{2}

Hence, we get Quotient = -x^{2} – 2

Remainder = -5x + 10

** **

**(2) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:**

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

**Solution: **

Given, first polynomial = t^{2} – 3

Second polynomial = 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

Hence, Remainder is zero. So the first polynomial is a factor of the second polynomial.

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

**Solution: **

Given, first polynomial = x^{2} + 3x + 1

Second polynomial = 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

Hence, remainder is zero. So the first polynomial is a factor of the second polynomial.

** **

**In case you are missed :- NCERT Solution for Pair Of Linear Equations In Two Variables**

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

**Solution: **

Given, First polynomial = x^{3} – 3x + 1

2^{nd }polynomial = x^{5} – 4x^{3} + x^{2 }+ 3x + 1

Hence, remainder is 2. So the first polynomial is not a factor of the second polynomial.

** **

**(3) Obtain all other zeroes of 3x ^{4} + 6x^{3} – 2x^{2} – 10x – 5, If two of its zeroes are **

**√5/3 and -√5/3.**

**Solution: **

3x^{4} + 6x^{3} – 2x^{2} – 10x – 5, If two of its zeroes are √5/3 and -√5/3

So, x = √5/3 and x = – √5/3

Or, (x – √5/3) = 0 or, (x + v5/3) = 0 –

**Now, (x – √5/3) (x + √5/3) = 0 **

(x)^{2} – (√5/3)^{2} = 0 [(a – b) (a + b) = a^{2} – b^{2}]

Or, x^{2} – 5/3 = 0

Or, 3x^{2} – 5 = 0

(3x^{2} – 5) = 0

(3x^{2} – 5) a factor of given polynomial

Hence, x^{2} + 2x + 1 is the other factor.

x^{2} + 2x + 1 = 0

x^{2} + x + x + 1 = 0

x (x + 1) +1 (x + 1) = 0

(x + 1) (x + 1)

x + 1 = 0

x = -1

x + 1 = 0

x = -1

Other zeroes are -1 and -1

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

**Solution: **

Dividend = Divisor × quotient + remainder

or divisor × quotient = dividend – remainder

or divisor = dividend – Remainder/quotient

∴ Hence g(x) = x^{2} – x + 1

** **

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

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

**Solution:**

__Example:-__

Let, p(x) = 2x^{2} – 2x + 2 is divided by 2

∴ p(x) = 2x^{2} – 2x + 2

q (x) = x^{2} – x + 2

deg p(x) = deg q(x) = 2

** **

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

**Solution: **

q (x) = x

r(x) = x

degq (x) = degr (x) = 1

** **

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

**Solution:**

So, degr9x) = 0

** **

__EXERCISE 2.4 (Optional)__

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

**Solution: **

p(x) = 2x^{3} + x^{2} – 5x + 2

p (1/2) = 2 × (1/2)^{3} + (1/2)^{2} – 5× ½ + 2

= 2 × 1/8 + 1/4 – 5/2 + 2

= 1/4 + 1/4 – 5/2 + 2

= 1+1-10+8/4 = 0/4 = 0

p(1) = 2× (1)^{3} + (1)^{2} -5×1 + 2

= 2 × 1 + 1 – 5 + 2

= 2 + 1 – 5 + 2 = 5 – 5 = 0

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

= 2 × (-8) + 4 + 10 + 2

= – 16 + 16 = 0

Here, α = – ½, β = 1, ð = -2

p(x) = 2x^{3} + x^{2} – 5x + 2,

∴ a = 2, b = 1, c = -5, d = 2

Now, α + β + ð = – b/a

1/2 + 1 – 2 = -1/2

– 1/2 = -1/2 (Verify)

Now, αβ + βð + ðα = c/a

½ × 1 + 1 × (-2) + (-2) × ½ = -5/2

Or, 1/2 – 2 – 1 = -5/2

Or, 1-4-2/2 = -5/2

Or, – 5/2 = – 5/2 (Verify)

α.βð = -d/a

½. 1 × (-2) = -2/2

-1 = -1 (Verify)

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

**Solution: **

p(x) = x^{3} – 4x^{2} + 5x – 2

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

= 8 – 16 + 10 – 2 = 18 – 18 = 0

p(1) = 1^{3} – 4 (1)^{2} + 5×1 – 2

= 1 – 4 + 5 – 2 = 6 – 6 = 0

p(1) = 1^{3} – 4(1)^{2} + 5×1 – 2

= 1 – 4 + 5 – 2 = 6 – 6 = 0

Here, α = 2, β = 1, ð = 1

p(x) = x^{3} – 4x^{2} + 5x – 2

∴ a = 1, b = -4, c = 5, d = -2

Now, α+β+ð = – b/a

Or, 2 + 1 + 1 = – (-4)/1

Or, 4 = 4 (verify)

αβ + βð + ðα = c/a

2×1 + 1×1 + 1×2 = 5/1

Or, 2 + 1 + 2 = 5

Or, 5 = 5 (Verify)

α.β.ð = – d/a

α.β.ð = – (-c)/d

Or, 2×1×1 = 2

Or, 2 = 2 Verify.

** **

**(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.**

**Solution: **

Let the cubic polynomial be

ax^{3} + bx^{2} + cx + d = 0

And the zeroes are α, β, ð.

∴ α + β + ð = – b/a = 2

∴ – a/b = 2

Or, b/a = -2/1

a = 1, b = -2

αβ + βð + ðα – c/a = -7

c/a = -7

Or, c/a = -7/1

C = -7 a = 1

α.β.ð = – d/a

∴ – d/a = -14

Or, d/a = 14/1

d = 14, a = 1

The cubic polynomial is

1.x^{3} + (-2) x^{2} + (-7) x + 14

= x^{3} – 2x^{2} – 7x + 14

** **

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

**Solution: **

p(x) = x^{3} – 3x^{2} + x + 1 and the zeroes are (a-b), a, (a+b)

p(x) = x^{3} – 3x^{2} + x + 1

a = 1, b = -3, c = 1, d = 1

α+β+ð = – b/a

a+b+a+a+b = – (-3)/1

3a = 3

a = 1

α.β.ð = – d/a

(a-b) × a × (a+b) = -1/1

a (a^{2} – b^{2}) = 1

1 (1^{2} – b^{2}) = 1

b^{2} = 1 + 1

b^{2} = 2

b = ± √2

Hence, (1 – √2), 1, (1 + √2) are the zeroes of x^{3} – 3x^{2} + x + 1.

**(4) If two zeroes of the polynomial x ^{4} – 6x^{3 }– 26x^{2} + 138x – 35 are 2 **

**±**

**√**

**3, find other zeroes.**

**Solution: **

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

So, x = 2 + √3 and x = 2 – √3

Or, x – (2 + √3) = 0 or, x – (2 – √3) = 0

Now, {x – (2 – √3)} (x – (2 – √3)}

= {(x – 2) – √3} {(x – 2) + √3)}

= {(x – 2)^{2} – (√3)^{2}}

= x^{2} – 4x + 4 = 3 = x^{2} – 4x + 1

So, x^{2} – 2x – 35 is the other factor.

x^{2} – 2x – 35 = 0

x^{2} – (7-5) x – 35 = 0

x^{2} – 7x – 5x – 35 = 0

x (x – 7) -5 (x – 7) = 0

(x – 5) (x – 7) = 0

x – 5 = 0

Or, x = 5

x – 7 = 0

x = 7

Hence, the other zeroes are 5 & 7.

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

**Solution: **

Remainder = (2K – 9) x + (10 – 8k + k^{2})

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

∴ Comparing this….

2K – 9 = 1

Or, 2k = 1+9

2K = 10

k = 5

10 – 8K + k^{2} = a

Or, a = 10 – 8.5 + 5^{2}

= 10 – 40 + 25

= 5

Hence, k = 5 and a = 5

**In case you are missed :- NCERT Solution for Quadratic Equations**