Factorise by splitting middle term
$4 x^{2} - x - 3$

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Solution

Let $p (x) = 4 x^{2} - x - 3$
Here we have to split middle term $- 1$ into two parts such that their sum is $- 1$ product is $3 \times - 4 = - 12$
$- 4$ and $3$ are such numbers whose sum is $- 1$ and product is $- 12$
$4 x^{2} - x - 3 = 4 x^{2} - 4 x + 3 x - 3$
$= 4 x (x - 1) + 3 (x - 1)$
$= (x - 1) (4 x + 3)$

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