Question

factorise???????????

Answer 1

it might be the answer

Answer 2

$( {x}^{2} - 4x)( {x}^{2} - 4x - 1) - 20$

Now,

Let y=x²-4x

Therefore,

$( {x}^{2} - 4x)( {x}^{2} - 4x - 1) - 20$

$=y(y - 1) - 20$

$= {y}^{2} - y - 20$

$= {y}^{2} - 5y + 4y - 20$

$= y(y - 5) + 4(y - 5)$

$= (y + 4)(y - 5)$

Now,

Putting the value of y, we get,

$( {x}^{2} - 4x + 4)( {x}^{2} - 4x - 5)$

$= ( {x}^{2} - 2x - 2x + 4)( {x}^{2} - 5x + x - 5)$

$= (x(x - 2) - 2(x - 2))(x(x - 5) + 1(x - 5))$

$= {(x - 2)}^{2} (x - 5)(x + 1)$

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