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.

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

**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

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.

**(iii)**Let p (x) = x

^{4}+ 3x

^{3}+ 3x

^{2}+ x + 1 .

Then, p (-1)= (-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 + √2) x + √2

Then, p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2

= -1 – 1+ 2 +√2+√2

= 2√2

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

x

^{3}– x

^{2}– (2 + √2) x + √2.

**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 + X 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).

**The zero of g (x) = x + 2 is – 2.**

(ii)

(ii)

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).

**The zero of g (x) = x – 3 is 3.**

(iii)

(iii)

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).

**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 + √2**(iii) p (x) = kx**

^{2}– √2 x + 1**(iv) p (x) = kx**

^{2}– 3x + k**Solution:**

The zero of x – 1 is 1.

**(i)**(x – 1) is a factor of p (x),then p(1)= 0 (By Factor theorem)

⇒ 1

^{2}+ 1 + k = 0 [∵ p(x) = x

^{2}+ x + k]

⇒ 2 + k =0

⇒ k = -2

**(ii)**∵ (x -1) is a factor of p (x), then p (1) = 0 (By Factor theorem)

⇒ 2(1)

^{2}+ k(1)+√2= 0 [∵p(x) = 2x

^{2}+ kx+ -√2]

⇒ 2 + k + √2 = 0

⇒ k = – (2 + √2)

**(iii)**∵ (x-1) is a factor of p (x), then p (1) = 0 (By Factor theorem)

⇒ k (1)

^{2}– √2 + 1 = 0 [∵p(x) = kx

^{2}– √2x + 1]

⇒ k = (√2 – 1)

**(iv)**∵ (x-1) is a factor of p (x), then p (1) = 0 (By Factor theorem)

⇒ k(1)

^{2}– 3 + k = 0 [∵p(x) = kx

^{2}– 3x + k]

⇒ 2k-3 = 0

⇒ k = \(\frac { 3 }{ 2 }\)

**Question 4.
**

**Factorise**

**(i) 12x**

^{2}– 7x +1**(ii) 2x**

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

^{2}+ 5x – 6**(iv) 3x**

^{2}– x – 4**Solution:**

**(i)**12x

^{2}– 7x + 1 = 12x

^{2}– 4x- 3x + 1 (Splitting middle term)

= 4x (3x – -0 -1 (3x-1)

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

**(ii)**2x

^{2}+ 7x + 3 = 2x

^{2}+ 6x + x + 3 (Splitting middle term)

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

**(iii)**6x

^{2}+ 5x – 6= 6x

^{2}+ 9x- 4x- 6 (Splitting middle term)

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

**(iv)**3x

^{2}– x- 4= 3x

^{2}-4x+3x-4 (Splitting middle term)

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

**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

p(1) = 0

By trial, we find that p (1) = 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) + 1 (x – 2)]

= (x – 1) (x – 2)(x + 1)

**Let p(x) = x**

(ii)

(ii)

^{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 }

**Let p (x) = x**

(iii)

(iii)

^{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+ 1) (x+ 10) (x + 2)

**Let p (y) = 2y**

(iv)

(iv)

^{3}+ y

^{2}– 2y -1

By trial we find that p(1) = 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) + 1 (y + 1)]

= (y – 1)(y+1)(2y+1)

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