Factorise :
$6 x^{2}+7 x-3$
Factorise :
$6 x^{2}+7 x-3$
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., $9 x-2 x$.
$\therefore$ $6 x^{2}+7 x-3=6 x^{2}+9 x-2 x-3$
$=3 x(2 x+3)-1(2 x+3)$
$=(2 x+3)(3 x-1)$
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