# How do you use synthetic division to find factors for x^3 + 2x^2 - 5x - 6?

Sep 26, 2015

Noting that $\left(x - 2\right)$ is a factor [see explanation], we can use synthetic division to reduce ${x}^{3} + 2 x - 5 x - 6$ to a more easily factored expression.
Complete factors: $\left(x - 2\right) \left(x + 3\right) \left(x + 1\right)$

#### Explanation:

Part 1: The initial Factor
If ${x}^{3} + 2 x - 5 x - 6 = 0$
then setting the negative terms on one side and the positive terms on the other:
$\textcolor{w h i t e}{\text{XX}} {x}^{3} + 2 {x}^{2} = 5 x + 6$
then testing a few values:

{: (x,color(white)("XX"),x^3+2x^2,color(white)("XX"),5x+6), (0,color(white)("XX"),0,color(white)("XX"),6), (1,color(white)("XX"),3,color(white)("XX"),11), (2,color(white)("XX"),16,color(white)("XX"),16) :}
So if $x = 2$ then ${x}^{3} + 2 {x}^{2} - 5 x - 6 = 0$
$\rightarrow \left(x - 2\right)$ is a factor of ${x}^{3} + 2 {x}^{2} - 5 x - 6$

Part 2: Use of synthetic division

So
$\textcolor{w h i t e}{\text{XX}} {x}^{2} + 2 {x}^{2} - 5 x - 6 = \left(x - 2\right) \left({x}^{2} + 4 x + 3\right)$

Part 3: Factoring the remaining quadratic
By observation or using the quadratic formula we can factor:
$\textcolor{w h i t e}{\text{XX}} {x}^{2} + 4 x + 3 = \left(x + 3\right) \left(x + 1\right)$

Part 4: Summarize results
${x}^{3} + 2 {x}^{2} - 5 x - 6 = \left(x - 2\right) \left(x + 3\right) \left(x + 1\right)$