# How do you factor x^3 -x^2 + x -6 = 0?

Jun 9, 2016

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

Derived by substitution and observation.

#### Explanation:

Given:$\text{ } {x}^{3} - {x}^{2} + x - 6$

Factors of 6 are {1,6} ; {2,3}
So $x$ equals at least one of these factors is a solution.
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Try $x = 1$
$1 - 1 + 1 - 6 \ne 0 \textcolor{red}{\text{ Fail}}$

Try $x = 2$
2^3-2^2+2-6=0 =>color(green)(x-2 " is a factor"
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So we have (x-2)(?)=x^3-x^2+x-6=0

Building by 1 term at a time.
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$\textcolor{b l u e}{\text{1st term:}}$ this needs to build ${x}^{3}$

(x-2)(x^2+?)  gives ${x}^{3}$

But $- 2 \times {x}^{2} = - 2 {x}^{2}$ giving

(x-2)(x^2+?) = x^3-2x^2+?

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$\textcolor{b l u e}{\text{2nd term}}$ this build to be $- {x}^{2}$

To 'force' $- 2 {x}^{2}$ into $- {x}^{2}$ we need to generate $+ {x}^{2}$

So our next number is $x$ giving

(x-2)(x^2+x+?) = x^3-2x^2+x^2-2x+?

(x-2)(x^2+x+?) = x^3-x^2-2x+?

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$\textcolor{b l u e}{\text{3rd term}}$ The $x$ from ${x}^{2} + x$ generated is $- 2 x$ but we need it to be $+ x$. So to force this change we need to generate $+ 3 x$

So our next number is 3 giving

(x-2)(x^2+x+3+?) =x^3-x^2-2x+3x-6 and this is our starting equation so the two bracket factors are:

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

We have now built all the terms

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The latter bracket can not be split into any further whole number factorisations so we stop.

The only whole number factors of 3 is {1,3} and $3 x - x \ne + x$