# Factorising Quartic Expressions with two Quadratic Factors and a Remainder, such as $(x^2+3x-2)(x^2+3x+4)-27$ $(x^2-8x+12)(x^2-7x+12)-6x^2$

## Example 1

Factorise $(x^2+3x-2)(x^2+3x+4)-27$.

\require{AMSsymbols} \begin{align} &= \left[ (\bbox[yellow]{x^2+3x})-2 \right] \left[ (\bbox[yellow]{x^2+3x})+4 \right]-27 \\ &= (\bbox[yellow]{x^2+3x})^2 + 2(\bbox[yellow]{x^2+3x})-8-27 \\ &= (\bbox[yellow]{x^2+3x})^2 + 2(\bbox[yellow]{x^2+3x})-35 \\ &= (\bbox[yellow]{x^2+3x}+7)(\bbox[yellow]{x^2+3x}-5) \end{align}

## Example 2

Factorise $(x^2-8x+12)(x^2-7x+12)-6x^2$.

\require{AMSsymbols} \begin{align} &= \left[ (x^2+12)-8x \right] \left[ (x^2+12)-7x \right]-6x^2 \\ &= (x^2+12)^2 -15x(x^2+12) + 56x^2-6x^2 \\ &= (x^2+12)^2\bbox[aqua]{-15x(x^2+12)} + 50x^2 \leadsto 50x^2 = \bbox[pink]{-10x} \times \bbox[yellow]{-5x} \end{align}

$\require{AMSsymbols} \begin{array} {ccr} &\bbox[yellow]{x^2+12} &\bbox[pink]{-10x} &\bbox[pink]{-10x(x^2+12)} \\ &\bbox[pink]{x^2+12} &\bbox[yellow]{-5x} &\bbox[yellow]{-5x(x^2+12)} \\ \hline &&&\bbox[aqua]{-15x(x^2+12)} \end{array}$

\require{AMSsymbols} \begin{align} &= (\bbox[yellow]{x^2+12} \bbox[pink,3px]{-10x})(\bbox[pink]{x^2+12}\bbox[yellow,3px]{-5x}) \\ &= (x^2-10x+12)(x^2-5x+12) \end{align}