NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4.

- Polynomials Class 9 Ex 2.1
- Polynomials Class 9 Ex 2.2
- Polynomials Class 9 Ex 2.3
- Polynomials Class 9 Ex 2.4
- Polynomials Class 9 Ex 2.5

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Polynomials |

Exercise |
Ex 2.4 |

Number of Questions Solved |
5 |

Category |
NCERT Solutions |

## NCERT Solutions 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} + 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 **+ **–\( \sqrt { 2 }\) )x + \( \sqrt { 2 }\)

**Solution:**

The zero of x +1 is -1.

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

Then, p(-1) = (-1)^{3} + (-1)^{2} + (-1) +1

= -1+1-1+1 => p(-1)= 0

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

**
(ii)** Let p(x)= x

^{4}+ x

^{3}+ x

^{2}+ x +1

Then, p(-1) = (-1)

^{4}+ (-1)

^{3}+ (-1)

^{2}+ (-1) +1

=1-1+1-1+1 => p(-1) = 1

So, by the.Factor theorem (x +1) is not a factor of x

^{4}+ x

^{3}+ x

^{2}+ x +1

**Let p(x) = x**

(iii)

(iii)

^{4}+ 3x

^{3}+ 3x

^{2}+ x +1

Then, p(-l) = (-1)

^{4}+ 3(-1)

^{3}+ 3(-1)

^{2}+ (-1) +1

=1-3+3-1+1

=> P(-1) = 1

So, by the Factor theorem (x +1) is not a factor of x

^{4}+ 3x

^{3}+ 3x

^{2}+ x +1

**(iv) **Let p(x) = x^{3}-x^{2}-(2 + \( \sqrt { 2 }\) )x + \( \sqrt { 2 }\)

Then, p(-1) = (-1)^{3} – (-1)^{2} – (2 + \( \sqrt { 2 }\) )(-1) + \( \sqrt { 2 }\)

= —1—1+2 + \( \sqrt { 2 }\) + \( \sqrt { 2 }\)

= 2\( \sqrt { 2 }\)

So, by the Factor theorem (x + 1) is not a factor ofx^{3} -x^{2} -(2 + \( \sqrt { 2 }\) x + \( \sqrt { 2 }\)

**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)** The zero of g(x) = x +1 is x = -1.

Then, p(-1)= 2(-1)^{3} +(-1)^{2} -2(-1)-1

[∵ p(x) = 2x^{3} + x^{2} -2x-1]

= -2 +1+2-1 ⇒ p(-1) = 0

Hence, g(x) is a factor of p(x).

**(ii)** The zero of g(x) = x + 2 is -2.

Then, p(-2) = (-2)^{3} + 3(-2)^{2} + 3(-2) +1

[∵ p(x) = x^{3} + 3x^{2} + 3x +1]

= -8+12-6 +1 =>p(-2)=-1

Hence, g(x) is not a factor of p(x).

**(iii)** The zero of g(x) = x – 3 is 3.

Then,p(3) = 3^{3} – 4(3)^{2} +3 + 6

[∵ p(x) = x^{3} – 4x^{2} + x + 6]

= 27-36 +3 + 6 =» p(3) = 0

Hence, g(x) is a factor of p(x).

**Ex 2.4 Class 9 Maths Question 3.**

Find the value of k, if x -1 is a factor of p(x) in each of the following cases:

(i) p(x) = x^{2}+x + k .

(ii) p(x) = 2x^{2} + kx + V2=

(iii) p(x) = kx^{2} -42x + 1

(iv) p(x) = kx^{2} -3x + k

**Solution:
**

**Ex 2.4 Class 9 Maths Question 4.**

**Factorise:**

(i) 12x

^{2}– 7x+1

(ii) 2x

^{2}+7x+3

(iii) 6x

^{2}+5x-6

(iv) 3x

^{2}-x-4

**Solution:**

**Ex 2.4 Class 9 Maths Question 5.**

**Factorise:**

(i) x

^{3}-2x

^{2}-x+2

(ii) x

^{3}-3x

^{2}-9x-5

(iii) x

^{3}-13x

^{2}+32x+20

(iv) 2y

^{3}+y

^{2}-2y-1

**Solution:**

**(i)**Let p(x) = x

^{3}-2x

^{2}-x+2, constant term of P(x) is 2.

Factors of 2 are ± 1 and ± 2.

Now, p(1)=1

^{3}-2(1)

^{2}-1+2

=1-2-1+2

By trial we find that p(l) = 0, so (x -1) is a factor of p(x).

So, x

^{3}-2x

^{2}-x+2= x

^{3}-x

^{2}-x

^{2}+ x-2x+2

= x

^{2}(x-1)-x(x-1)-2(x-1)

= (x—1)(x

^{2}-x-2)

= (x -1)(x

^{2}-2x + x -2)

= (x -1) [x(x -2) + l(x -2)]

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

**(ii)** Let p(x)= x^{3} -3x^{2} -9x-5

By trial, we find that p(5) = (5)^{3} – 3(5)^{2} – 9(5) – 5

= 125 – 75 – 45 – 5 = 0

So, (x – 5) is a factor of p(x).

So, x^{3}-3x^{2}-9x-5=x^{3}-5x^{2}+2x^{2} -10x + x – 5

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

= (x – 5) (x^{2}+2x+1)

= (x – 5) (x^{2} + x + x +1)

= (x + 5) [x(x +1) + 1(x +1)]

= (x-5)(x +1) (x +1)

= (x – 5)(x +1)^{2}

**(iii)** Let p(x)= x^{3} +13x^{2} +32x +20

By trial, we find that p(-1) = (-1)^{3} +13(-1)^{2} + 32(-1) + 20

= -1 +13 -32 + 20 = -33 + 33 = 0

So (x +1) is a factor of p(x).

So, x^{3} + 13x^{2} + 32x + 20 = x^{3} + x^{2} + 12x^{2} + 12x + 20x + 20

= x^{2} (x +1) + 12x(x +1) + 20(x +1) = (x+1)(x^{2} + 12x + 20)

= (x+1)(x^{2} + 10x + 2x + 20)

= (x +1)[x(x +10) + 2(x +10)]

= (x +l)(x +10)(x + 2)

**(iv)** Let p(y) = 2y^{3} + y^{2} – 2y -1

By trial we find that p(l) = 2(1)^{3} + (1)^{2} -2(1) -1,

=2+1-2-1=0 So (y -1) is a factor of p(y).

So, 2y^{3} + y^{2} -2y-1 = 2y^{3} -2y^{2} +3y^{2} -3y + y-1

= 2y^{2}(y-1) + 3y(y-1) + 1(y-1)

= (y – 1)(2y^{2} +3y + 1)

= (y – 1)(2y^{2} + 2y + y + 1)

= (y -1) [2y(y +1) + l(y +1)

= (y-1)(y+ l)(2y+ 1)

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