# NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1 are part of NCERT Solutions for Class 9 Maths . Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1.

- Polynomials
- Introduction
- Polynomials In One Variable
- Zeroes Of A Polynomial
- Remainder Theorem
- Factorisation Of Polynomials
- Algebraic Identities
- Summary

## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1

### Ex 2.1 Class 9 Maths Question 1.

Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.(i) 4x

^{2}– 3x + 7

(ii) y

^{2}+ √2

(iii) 3 √t + t√2

(iv) y+2/y

(v) x

^{10}+ y

^{3}+t

^{50}

**Solution:**

(i) We have 4x

^{2}– 3x + 7 = 4x

^{2}– 3x + 7x

^{0}

It is a polynomial in one variable i.e., x

because each exponent of x is a whole number.

(ii) We have y^{2} + √2 = y^{2} + √2y^{0}

It is a polynomial in one variable i.e., y

because each exponent of y is a whole number.

(v) We have x^{10}+ y^{3 }+ t^{50}

Here, exponent of every variable is a whole number, but x^{10} + y^{3} + t^{50} is a polynomial in x, y and t, i.e., in three variables.

So, it is not a polynomial in one variable.

Ex 2.1 Class 9 Maths Question 2.

Write the coefficients of x^{2} in each of the following

(i) 2 + x^{2} + x

(ii) 2 – x^{2} + x^{3}

(iii)π/2 x^{2} + x

(iv) v2 x – 1

**Solution:**

(i) The given polynomial is 2 + x^{2} + x.

The coefficient of x^{2} is 1.

(ii) The given polynomial is 2 – x^{2} + x^{3}.

The coefficient of x^{2} is -1.

(iii) The given polynomial is π/2 x^{2}+ x.

The coefficient of x^{2} is π/2.

(iv) The given polynomial is v2 x – 1.

The coefficient of x^{2} is 0.

Ex 2.1 Class 9 Maths Question 3.

Give one example each of a binomial of degree 35, and of a monomial of degree 100.

**Solution:**

(i) Abmomial of degree 35 can be 3x^{35} -4.

(ii) A monomial of degree 100 can be √2y^{100}.

## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

### Ex 2.2 Class 9 Maths Question 1.

Find the value of the polynomial 5x – 4x^{2} + 3 at

(i) x = 0

(ii) x = – 1

(iii) x = 2

**Solution:**

1et p(x) = 5x – 4x^{2} + 3

(i) p(0) = 5(0) – 4(0)^{2} + 3 = 0 – 0 + 3 = 3

Thus, the value of 5x – 4x^{2} + 3 at x = 0 is 3.

(ii) p(-1) = 5(-1) – 4(-1)^{2} + 3

= – 5x – 4x^{2} + 3 = -9 + 3 = -6

Thus, the value of 5x – 4x^{2} + 3 at x = -1 is -6.

(iii) p(2) = 5(2) – 4(2)^{2} + 3 = 10 – 4(4) + 3

= 10 – 16 + 3 = -3

Thus, the value of 5x – 4x^{2} + 3 at x = 2 is – 3.

### Ex 2.2 Class 9 Maths Question 2.

Find p (0), p (1) and p (2) for each of the following polynomials.

(i) p(y) = y^{2} – y +1

(ii) p (t) = 2 +1 + 2t^{2} -t^{3 }

(iii) P (x) = x^{3}

(iv) p (x) = (x-1) (x+1)

**Solution:**

(i) Given that p(y) = y^{2} – y + 1.

∴ P(0) = (0)^{2} – 0 + 1 = 0 – 0 + 1 = 1

p(1) = (1)^{2} – 1 + 1 = 1 – 1 + 1 = 1

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

(ii) Given that p(t) = 2 + t + 2t^{2 } – t^{3 }

∴p(0) = 2 + 0 + 2(0)^{2 } – (0)^{3 }

= 2 + 0 + 0 – 0=2

P(1) = 2 + 1 + 2(1)^{2 } – (1)^{3}

= 2 + 1 + 2 – 1 = 4

p( 2) = 2 + 2 + 2(2)^{2 } – (2)^{3}

= 2 + 2 + 8 – 8 = 4

(iii) Given that p(x) = x^{3}

∴ p(0) = (0)^{3} = 0, p(1) = (1)^{3} = 1

p(2) = (2)^{3} = 8

(iv) Given that p(x) = (x – 1)(x + 1)

∴ p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1

p(1) = (1 – 1)(1 +1) = (0)(2) = 0

P(2) = (2 – 1)(2 + 1) = (1)(3) = 3

## NCERT So1utions for C1ass 9 Maths Chapter 2 Polynomials Ex 2.3

### Ex 2.3 Class 9 Maths Question 1.

Find the remainder when x^{3}+ 3x

^{2}+ 3x + 1 is divided by

(i) x + 1

(ii) x – 1/2

(iii) x

(iv) x + π

(v) 5 + 2x

**Solution:**

Let p(x) = x

^{3}+ 3x

^{2}+ 3x +1

(i) The zero of x + 1 is -1.

∴ p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1

= -1 + 3- 3 + 1 = 0

Thus, the required remainder = 0

### Ex 2.3 Class 9 Maths Question 2.

Find the remainder when x^{3} – ax^{2} + 6x – a is divided by x – a.

**Solution:**

We have, p(x) = x^{3} – ax^{2} + 6x – a and zero of x – a is a.

∴ p(a) = (a)^{3} – a(a)^{2} + 6(a) – a

= a^{3} – a^{3} + 6a – a = 5a

Thus, the required remainder is 5a.

## NCERT So1utions for Class 9 Maths Chapter 2 Polynomials Ex 2.4

### Ex 2.4 Class 9 Maths Question 1.

Determine which of the following polynomials has (x +1) a factor.

(i) x^{3}+x^{2}+x +1

(ii) x^{4} + x^{3} + x^{2} + x + 1

(iii) x^{4} + 3x^{3} + 3x^{2} + x + 1

(iv) x^{3} – x^{2} – (2 +√2 )x + √2

**Solution:**

The zero of x + 1 is -1.

(i) Let p (x) = x^{3} + x^{2} + x + 1

∴ p (-1) = (-1)^{3} + (-1)^{2} + (-1) + 1 .

= -1 + 1 – 1 + 1

⇒ p (- 1) = 0

So, (x+ 1) is a factor of x^{3} + x^{2} + x + 1.

### Ex 2.4 Class 9 Maths Question 2.

Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases

(i) p (x)= 2x^{3} + x^{2} – 2x – 1, g (x) = x + 1

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

(iii) p (x) = x^{3} – 4x^{2} + x + 6, g (x) = x – 3

**Solution:**

(i) We have, p (x)= 2x^{3} + x^{2} – 2x – 1 and g (x) = x + 1

∴ p(-1) = 2(-1)^{3} + (-1)^{2} – 2(-1) – 1

= 2(-1) + 1 + 2 – 1

= -2 + 1 + 2 -1 = 0

⇒ p(-1) = 0, so g(x) is a factor of p(x).

## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5

### Ex 2.5 Class 9 Maths Question 1.

Use suitable identities to find the following products

(i) (x + 4)(x + 10)

(ii) (x+8) (x -10)

(iii) (3x + 4) (3x – 5)

### Ex 2.5 Class 9 Maths Question 2.

Evaluate the following products without multiplying directly

(i) 103 x 107

(ii) 95 x 96

(iii) 104 x 96

**Solution:**

(i)We have, 103 x 107 = (100 + 3) (100 + 7)

= ( 100)^{2} + (3 + 7) (100)+ (3 x 7)

[Using (x + a)(x + b) = x^{2} + (a + b)x + ab]

= 10000 + (10) x 100 + 21

= 10000 + 1000 + 21=11021

(ii) We have, 95 x 96 = (100 – 5) (100 – 4)

= ( 100)^{2} + [(- 5) + (- 4)] 100 + (- 5 x – 4)

[Using (x + a)(x + b) = x^{2} + (a + b)x + ab]

= 10000 + (-9) + 20 = 9120

= 10000 + (-900) + 20 = 9120

(iii) We have 104 x 96 = (100 + 4) (100 – 4)

= (100)^{2}-4^{2 }

[Using (a + b)(a -b) = a^{2}– b^{2}]

= 10000 – 16 = 9984