Factorise:
(i) $12 x^{2} - 7 x + 1$
(ii) $2 x^{2} + 7 x + 3$

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Solution

(i) $12 x^{2} - 7 x + 1 = 12 x^{2} - 4 x - 3 x + 1 = 4 x (3 x - 1) - 1 (3 x - 1) = (4 x - 1) (3 x - 1)$
(ii) $2 x^{2} + 7 x + 3 = 2 x^{2} + 6 x + x + 3 = 2 x (x + 3) + 1 (x + 3) = (2 x + 1) (x + 3)$

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Similar questions

Factorise:
(i) $12 x^{2} - 7 x + 1$
(ii) $2 x^{2} + 7 x + 3$
(iii) $6 x^{2} + 5 x - 6$
(iv) $3 x^{2} - x - 4$

12 x^{2} - 7 x + 1

2 x^{2} + 7 x + 3

6 x^{2} + 5 x - 6

3 x^{2} - x - 4

\frac{x^{2} + 7 x + 12}{x^{2} + 4 x + 3}

Factorise: $12 x^{2} - 7 x + 1$

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