# How do you use synthetic division to find all roots of x^3 + 5x^2 + 3x -9 = 0?

May 25, 2018

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

$x = 1 , \textcolor{w h i t e}{\text{dd}} \underbrace{x = - 3 \mathmr{and} x = - 3}$
$\textcolor{w h i t e}{\text{dddddddddddddd.d}} \uparrow$
color(white)("dddddddddddd")" duplicity"

#### Explanation:

Whole number factors of 9 are ${3}^{2} \mathmr{and} 1 \times 9$

Lets test for $x = + 1$

${\left(+ 1\right)}^{3} + 5 {\left(+ 1\right)}^{2} + 3 \left(+ 1\right) - 9$
$\textcolor{w h i t e}{\text{dd")1color(white)("dd.")+color(white)("dd")5color(white)("dd.d")+color(white)("dd")3color(white)("dd.}} - 9 = 0$ as required

So we have $\left(x - 1\right)$ as a factor.

To determine the related quadratic divide by $\left(x - 1\right)$

$\frac{{x}^{3} + 5 {x}^{2} + 3 x - 9}{x - 1} = 0 = \text{ some quadratic}$

The coefficients for ${x}^{3} + 5 {x}^{2} + 3 x - 9 \textcolor{w h i t e}{\text{dd")->color(white)("dd}} 1 , 5 , 3 , - 9$

color(white)("d")1|1color(white)("d")5color(white)("d")3color(white)("d")-9 color(white)("d")larr" The coefficients"
color(white)("dd.")ul(|darr1color(white)("d")6color(white)("ddd.")9)
$\textcolor{w h i t e}{\text{dddd")1 color(white)("d") 6color(white)("d")9color(white)("ddd")0color(white)("d")->color(white)("d}} 1 {x}^{2} + 6 x + 9$

So we have:

$\left(x - 1\right) \left({x}^{2} + 6 x + 9\right) = 0$

Note that $3 \times 3 = 9 \mathmr{and} 3 + 3 = 6$

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