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

Jan 17, 2017

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

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

Putting $x = - 1$ we get

$y = {\left(- 1\right)}^{3} - 4 {\left(- 1\right)}^{2} + \left(- 1\right) + 6$
$= - 1 - 4 - 1 + 6 = 0$

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

$y = {x}^{3} - 4 {x}^{2} + x + 6$
$= {x}^{3} + {x}^{2} - 5 {x}^{2} - 5 x + 6 x + 6$
$= {x}^{2} \left(x + 1\right) - 5 x \left(x + 1\right) + 6 \left(x + 1\right)$

$= \left(x + 1\right) \left({x}^{2} - 5 x + 6\right)$

$= \left(x + 1\right) \left({x}^{2} - 3 x - 2 x + 6\right)$

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

$= \left(x + 1\right) \left(x - 3\right) \left(x - 2\right)$