Question
Factorize the following polynomials: $\left(x^2-6 x\right)^2-8\left(x^2-6 x+8\right)-64$
✓
Answer
$\left(x^2-6 x\right)^2-8\left(x^2-6 x+8\right)-64$
$=m^2-8(m+8)-64 \ldots\left[\text { Putting } x^2-6 x=m\right]$
$=m^2-8 m-64-64$
$=m^2-8 m-128$
$=m^2-16 m+8 m-128$
$=m(m-16)+8(m-16)$
$=(m-16)(m+8)$
$=\left(x^2-6 x-16\right)\left(x^2-6 x+8\right) \ldots\left[\text { Replacing } m=x^2-6 x\right]$
$=\left(x^2-8 x+2 x-16\right)\left(x^2-4 x-2 x+8\right)$
$=[x(x-8)+2(x-8)][x(x-4)-2(x-4)]$
$=(x-8)(x+2)(x-4)(x-2)$
$=m^2-8(m+8)-64 \ldots\left[\text { Putting } x^2-6 x=m\right]$
$=m^2-8 m-64-64$
$=m^2-8 m-128$
$=m^2-16 m+8 m-128$
$=m(m-16)+8(m-16)$
$=(m-16)(m+8)$
$=\left(x^2-6 x-16\right)\left(x^2-6 x+8\right) \ldots\left[\text { Replacing } m=x^2-6 x\right]$
$=\left(x^2-8 x+2 x-16\right)\left(x^2-4 x-2 x+8\right)$
$=[x(x-8)+2(x-8)][x(x-4)-2(x-4)]$
$=(x-8)(x+2)(x-4)(x-2)$
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