# How do you use synthetic substitution to find the real zeroes for P(x)=x^4-6x^3+11x^2-8x+2?

Aug 5, 2015

The real zeroes of $P \left(x\right) = {x}^{4} - 6 {x}^{3} + 11 {x}^{2} - 8 x + 2$ are $\textcolor{red}{1}$ and $\textcolor{red}{1}$ and $\textcolor{red}{2 + \sqrt{2}}$ and $\textcolor{red}{2 - \sqrt{2}}$.

#### Explanation:

According to the rational root theorem, the rational roots of $f \left(x\right) = 0$ must all be of the form $\frac{p}{q}$ with $p$ a divisor of $2$ and $q$ a divisor of $1$.

So the only possible rational roots are $1$ and $2$.

We have to test both possibilities.

The only one that works is.

$1 | 1 \text{ "-6" "11" " "-8" " } \textcolor{w h i t e}{1} 2$
$\textcolor{w h i t e}{1} | \text{ " " "color(white)(1)1color(white)(1)-5" } \textcolor{w h i t e}{1} 6 - 2$
" "stackrel("————————————————————)
$\text{ "color(white)(1)1" "-5" "color(white)(1)6color(white)(1)-2 " " } \textcolor{red}{0}$

So $1$ is a real zero of the polynomial.

That means that $x - 1$ and ${x}^{3} - 5 {x}^{2} + 6 x - 2$ are factors of $P \left(x\right)$.

A fourth degree polynomial must have an even number of real zeroes.

The real zeroes of ${x}^{3} - 5 {x}^{2} + 6 x - 2$ must all be of the form $\frac{p}{q}$ with $p$ a divisor of $- 2$ and $q$ a divisor of $1$.

The possible real zeroes are ±1 and ±2.

The only one that works is

$1 | 1 \text{ "-5" "color(white)(1)6" " } - 2$
$\textcolor{w h i t e}{1} | \text{ " " "color(white)(1)1color(white)(1)-4" " " } 2$
" "stackrel("————————————————————)
$\text{ "color(white)(1)1" "-4" "color(white)(1)2" " " } \textcolor{red}{0}$

So $1$ is a second real zero.

The remaining factor is ${x}^{2} - 4 x + 2$.

Whose zeros can be found by using the quadratic formula.

${x}^{2} - 4 x + 2 = 0$

$x = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(1\right) \left(2\right)}}{2 \left(1\right)}$

$= \frac{4 \pm \sqrt{8}}{2}$

$= 2 \pm \sqrt{2}$

The real zeroes of $P \left(x\right)$ are $1$ and $1$ and $2 + \sqrt{2}$ and $2 - \sqrt{2}$.