Factorise :
$6 x^{2} + 7 x - 3$

Open in App

Solution

In order to factorise $6 x^{2} + 7 x - 3$ , we have to find two numbers p and q such that $p + q = 7$ and $p q = - 18$ .

Clearly, $9 + (- 2) = 7$ and $9 \times (- 2) = - 18$ .

So, we write the middle term $7 x$ as $9 x + (- 2 x),$ i.e., $9 x - 2 x$ .

$∴ 6 x^{2} + 7 x - 3 = 6 x^{2} + 9 x - 2 x - 3$

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

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

Suggest Corrections

Similar questions

Factorise : $5 + 7 x - 6 x^{2}$

x^{2} + 9 x + 18

6 x^{2} + 7 x - 3

2 x^{2} - 7 x - 15

84 - 2 r - 2 r^{2}

Tick $(✓)$ the correct answer in each of the following:
$(2 x + 3) (3 x - 1) = ?$

(a) $(6 x^{2} + 8 x - 3)$
(b) $(6 x^{2} + 7 x - 3)$
(c) $6 x^{2} - 7 x - 3$
(d) $(6 x^{2} - 7 x + 3)$

6 x^{2} - 7 x - 3

Join BYJU'S Learning Program