# How do you factor x^3-3x^2+x-55?

Oct 7, 2015

$\left(x - 5\right) \left({x}^{2} + 2 x + 11\right)$

#### Explanation:

If ${x}^{3} - 3 {x}^{2} + x - 55$ has an integer root $r$ then
$r$ must be a factor of $55$
i.e. $r \in \left\{\pm 5 , \pm 11\right\}$

A quick check reveals $r = 5$ satisfies the requirement
so
$\left(x - r\right) = \left(x - 5\right)$ is a factor

Dividing ${x}^{3} - 3 {x}^{2} + x - 55$ by $\left(x - 5\right)$
$\textcolor{w h i t e}{\text{XXX}}$(for example by using synthetic division:)

{:(," | ", 1, -3,color(white)("X")1, -55), (+5," | ",,color(white)("X")5,10,color(white)("X")55),(,"---","---","---","---","----"),(,,color(red)(1),color(red)(2),color(red)(11),color(white)("X")color(green)(0)):}

gives
$\left(x - 5\right) \left(\textcolor{red}{1} {x}^{2} + \textcolor{red}{2} x + \textcolor{red}{11}\right)$

Checking the discriminant of $\left({x}^{2} + 2 x + 11\right)$ reveals that no further Real factoring is possible.