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

Solution:

(i) $12 x^{2}-7 x+1$

We can find two numbers such that $p q=12 \times 1=12$ and $p+q=-7$. They are $p=-4$ and $q=-3$.

Here, $12 x^{2}-7 x+1=12 x^{2}-4 x-3 x+1$

$=4 x(3 x-1)-1(3 x-1)$

$=(3 x-1)(4 x-1)$

(ii) $2 x^{2}+7 x+3$

We can find two numbers such that $p q=2 \times 3=6$ and $p+q=7$.

They are $p=6$ and $q=1$.

Here, $2 x^{2}+7 x+3=2 x^{2}+6 x+x+3$

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

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

(iii) $6 x^{2}+5 x-6$

We can find two numbers such that $p q=-36$ and $p+q=5$.

They are $p=9$ and $q=-4$.

Here,

$6 x^{2}+5 x-6=6 x^{2}+9 x-4 x-6$

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

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

(iv) $3 x^{2}-x-4$

We can find two numbers such that $p q=3 \times(-4)=-12$

and $p+q=-1$.

They are $p=-4$ and $q=3$.

Here,

$3 x^{2}-x-4=3 x^{2}-4 x+3 x-4$

$=x(3 x-4)+1(3 x-4)$

$=(3 x-4)(x+1)$

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