Can $y=x^3-4x^2+x-4+10$ be factored? If so what are the factors ?

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Answer 1 · 2017-01-17T14:04:04

$y=x^3-4x^2+x-4+10$

$y=x^3-4x^2+x+6$

Putting $x=-1$ we get

$y=(-1)^3-4(-1)^2+(-1)+6$
$=-1-4-1+6=0$

So ( $x+1$ ) is a factor of the polynomial of x

$y=x^3-4x^2+x+6$
$=x^3+x^2-5x^2-5x+6x+6$
$=x^2(x+1)-5x(x+1)+6(x+1)$

$=(x+1)(x^2-5x+6)$

$=(x+1)(x^2-3x-2x+6)$

$=(x+1){x(x-3)-2(x-3)}$

$=(x+1)(x-3)(x-2)$

Can y=x^3-4x^2+x-4+10 y=x^3-4x^2+x-4+10 be factored? If so what are the factors ?