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

(i) Using splitting method, split the middle term as follows:

$\begin{array}{rcl} & =& 12x^2-7x+1\\ & =& 12x^2-4x-3x+1\end{array}$

Splitting the term in such a way that the product of two constants must be equal to coefficient of $x^2$, i.e. $-4\times -3=12$ and the sum of two constants must be equal to coefficient of $x$, i.e. $-4+\left(-3\right)=-7$

Hence, we get:

$\begin{array}{rcl} & =& 4x\left(3x-1\right)-1\left(3x-1\right)\\ & =& \left(4x-1\right)\left(3x-1\right)\end{array}$

(ii) Using splitting method, split the middle term as follows:

$\begin{array}{rcl} & =& 2x^2+7x+3\\ & =& 2x^2+6x+x+3\end{array}$

Splitting the term in such a way that the product of two constants must be equal to coefficient of $x^2$, i.e. $6\times 1=6$ and the sum of two constants must be equal to coefficient of $x$, i.e. $6+1=7$

Hence, we get:

$\begin{array}{rcl} & =& 2x\left(x+3\right)+1\left(x+3\right)\\ & =& \left(2x+1\right)\left(x+3\right)\end{array}$