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Question

Factorise:

(i) 12x2 − 7x + 1 (ii) 2x2 + 7x + 3

(iii) 6x2 + 5x − 6 (iv) 3x2x − 4

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Solution

(i) 12x2 − 7x + 1

We can find two numbers such that pq = 12 × 1 = 12 and p + q = −7. They are p = −4 and q = −3.

Here, 12x2 − 7x + 1 = 12x2 − 4x − 3x + 1

= 4x (3x − 1) − 1 (3x − 1)

= (3x − 1) (4x − 1)

(ii) 2x2 + 7x + 3

We can find two numbers such that pq = 2 × 3 = 6 and p + q = 7.

They are p = 6 and q = 1.

Here, 2x2 + 7x + 3 = 2x2 + 6x + x + 3

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

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

(iii) 6x2 + 5x − 6

We can find two numbers such that pq = −36 and p + q = 5.

They are p = 9 and q = −4.

Here,

6x2 + 5x − 6 = 6x2 + 9x − 4x − 6

= 3x (2x + 3) − 2 (2x + 3)

= (2x + 3) (3x − 2)

(iv) 3x2x − 4

We can find two numbers such that pq = 3 × (− 4) = −12

and p + q = −1.

They are p = −4 and q = 3.

Here,

3x2x − 4 = 3x2 − 4x + 3x − 4

= x (3x − 4) + 1 (3x − 4)

= (3x − 4) (x + 1)



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