Question
Factorise : $\left(y^2-3 y\right)^2-5\left(y^2-3 y\right)-50$.

Answer

Let $\left(y^2-3 y\right)=x$
$\therefore\left(y^2-3 y\right) 2-5\left(y^2-3 y\right)-50=x^2-5 x-50$
$=x^2-10 x+5 x-50$
$=x(x-10)+5(x-10)$
$=(x-10)(x+5)$
$=\left(y^2-3 y-10\right)\left(y^2-3 y+5\right)$
$=\left[y^2-5 y+2 y-10\right]\left(y^2-3 y+5\right)$
$=[y(y-5)+2(y-5)]\left(y^2-3 y+5\right)$
$=(y-5)(y+2)\left(y^2-3 y+5\right)$

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