Question
Factorise: $(x+2)(x-3)(x-7)(x-2)+64$
✓
Answer
$(x+2)(x-3)(x-7)(x-2)+64$
$=(x+2)(x-7)(x-3)(x-2)+64$
$=\left(x^2-5 x-14\right)\left(x^2-5 x+6\right)+64$
$=(m-14)(m+6)+64 \ldots \ldots \ldots\left(\text { putting } x^2-5 x=m\right)$
$=m^2-14 m+6 m-84+64$
$=m^2-8 m-20$
$=(m-10)(m+2)$
$=\left(x^2-5 x-10\right)\left(x^2-5 x+2\right) \ldots .\left(\text { replace } m \text { with } x^2-5 x\right)$
$=(x+2)(x-7)(x-3)(x-2)+64$
$=\left(x^2-5 x-14\right)\left(x^2-5 x+6\right)+64$
$=(m-14)(m+6)+64 \ldots \ldots \ldots\left(\text { putting } x^2-5 x=m\right)$
$=m^2-14 m+6 m-84+64$
$=m^2-8 m-20$
$=(m-10)(m+2)$
$=\left(x^2-5 x-10\right)\left(x^2-5 x+2\right) \ldots .\left(\text { replace } m \text { with } x^2-5 x\right)$
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