How do you factor $x^2-8x-48$ ?

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Answer 1 · 2016-04-05T16:12:49

$x^2-8x-48=(x+4)(x-12)$

To factorize a polynomial of general form $ax^2+bx+c$ , one should divide middle term $bx$ in two parts, whose sum is $b$ and product is $axxc$ .

So for $x^2-8x-48$ , we should divide $-8x$ in two parts which add up to $-8$ and whose product is $1xx(-48)=-48$ .

A few trial indicate them to be $-12$ and $4$ . Hence

$x^2-8x-48=x^2-12x+4x-48$ or

$x(x-12)+4(x-12)$ or

$(x+4)(x-12)$

How do you factor x^2-8x-48x^2-8x-48?