Question

"Question 4 Factorise: (i) 12x^2 − 7x + 1 (ii) 2x^2 + 7x + 3 (iii) 6x^2 + 5x − 6 (iv) 3x^2 − x − 4

Class 9 - Math - Polynomials Page 44"

Answer 1

Here we factorise it by splitting the middle term.

 

To factorise quadratic polynomial of the type ax²+bx+c by middle term splitting ,write b as the sum of two numbers whose product is ac.

 

 

To factorise ax²+bx- c & ax²-bx-c , write b as the difference of two numbers whose product is -ac.

 

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Solution:

 

 

(i) 12x² + 7x + 1
= 12x² – 4x – 3x+ 1
= 4x(3x–1)–1(3x – 1)
= (3x – 1) (4x – 1)

 

 

(ii) 2x² + 7x + 3
= 2x² + 6x + x + 3
= 2x (x + 3) +1(x + 3)
=  (x + 3) (2x + 1)

(iii) 6x² + 5x – 6
= 6x² + 9x – 4x – 6

=3x(2x + 3)–2(2x+ 3)
= (2x + 3) (3x – 2)

(iv) 3x² – x – 4
= 3x² – 4x + 3x – 4
= x(3x – 4)+1(3x – 4)
= (3x – 4) (x + 1)

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Answer 2

i)12x2-7x+1

12x2-(4+3)x+1=0
12x2-4x-3x+1=0
4x (3x-1)-1 (3x-1)=0
(3x-1)(4x-1)=0 , x = 1/3 or x=1/4

ii)2x2+7x+3=0
2x2+(6+1)x+3=0
2x2+6x+x+3=0
2x (x+3)+1 (x+3)=0
(2x+1)(x+3)=) , x=-1/2 or x=-3

iv)3x2-x-4=0
3x2-(4-3)x-4=0
3x2-4x+3x-4=0
x (3x-4)+1 (3x-4)=0
(3x-4)(x+1)=0 , x=-1, x=4/3

iii)6x2+5x-6=0
6x2+(9-4)x-6=0
6x2+9x-4x-6=0
3x (2x+3)-2 (2x+3)=0
(2x+3)(3x-2)=0 , x=-3/2 or x=2/3

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