How do you use synthetic division to show that x=1+sqrt3 is a zero of x^3-3x^2+2=0?

Aug 1, 2018

Remainder $= 0 \implies x = 1 + \sqrt{3}$ is a zero of $\left({x}^{3} - 3 {x}^{2} + 2\right)$

Explanation:

Using synthetic division :

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

We have , $p \left(x\right) = {x}^{3} - 3 {x}^{2} + 0 x + 2 \mathmr{and} \text{divisor :} x = 1 + \sqrt{3}$

We take ,coefficients of $p \left(x\right) \to 1 , - 3 , 0 , 2$

$1 + \sqrt{3} |$ $1 \textcolor{w h i t e}{\ldots \ldots \ldots . .} - 3 \textcolor{w h i t e}{\ldots \ldots \ldots . .} 0 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots} 2$
$\underline{\textcolor{w h i t e}{\ldots \ldots \ldots .}} |$ ul(0color(white)( .......)1+sqrt3color(white)(.......)1-sqrt3color(white)(......)-2
color(white)(.............)1color(white)(...)-2+sqrt3color(white)(.....)1-sqrt3color(white)(........)color(violet)(ul|0|
We can see that , quotient polynomial :

$q \left(x\right) = {x}^{2} + \left(- 2 + \sqrt{3}\right) x + 1 - \sqrt{3} \mathmr{and} \text{the Remainder} = 0$

Hence ,

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

Remainder $= 0 \implies x = 1 + \sqrt{3}$ is a zero of $\left({x}^{3} - 3 {x}^{2} + 2\right)$