Question:

Factorise : (i) 12x 2 – 7x + 1 (ii) 2x 2 + 7x + 3 (iii) 6x 2 + 5x – 6 (iv) 3x 2 – x – 4.

Updated On: Jun 21, 2024
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Solution and Explanation

(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) 3x2 − x − 4 We can find two numbers such that pq = 3 × (− 4) = −12 and p + q = −1. 

They are p = −4 and q = 3. 

Here, 3x2 − x − 4 = 3x2 − 4x + 3x − 4 

= x (3x − 4) + 1 (3x − 4) = (3x − 4) (x + 1).

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