# How do you use polynomial synthetic division to divide (x^4-6x^2+9)div(x-sqrt3) and write the polynomial in the form p(x)=d(x)q(x)+r(x)?

Aug 12, 2018

$\frac{{x}^{4} - 6 {x}^{2} + 9}{x - \sqrt{3}} = \left(x - \sqrt{3}\right) \left({x}^{2} + 2 \sqrt{3} x + 3\right) + 0$

#### Explanation:

$\frac{{x}^{4} - 6 {x}^{2} + 9}{x - \sqrt{3}}$

$= \frac{{x}^{3} \left(x - \sqrt{3}\right) + \sqrt{3} {x}^{3} - 6 {x}^{2} + 9}{x - \sqrt{3}}$

$= \frac{{x}^{3} \left(x - \sqrt{3}\right) + \sqrt{3} {x}^{2} \left(x - \sqrt{3}\right) - 3 {x}^{2} + 9}{x - \sqrt{3}}$

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

$= {x}^{3} + \sqrt{3} {x}^{2} - 3 x - 3 \sqrt{3} + 0$

$\sqrt{3}$ is the solution again, so $x - \sqrt{3}$ divide this polynomial.

$\frac{{x}^{3} + \sqrt{3} {x}^{2} - 3 x - 3 \sqrt{3}}{x - \sqrt{3}}$

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

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

$= {x}^{2} + 2 \sqrt{3} x + 3$

So:

$\frac{{x}^{4} - 6 {x}^{2} + 9}{x - \sqrt{3}} = \left(x - \sqrt{3}\right) \left({x}^{2} + 2 \sqrt{3} x + 3\right) + 0$

( and we can notice that $\sqrt{3}$ is a double solution of ${x}^{4} - 6 {x}^{2} + 9$)

Aug 12, 2018

$\left({x}^{4} - 6 {x}^{2} + 9\right) = \left(x - \sqrt{3}\right) \left({x}^{3} + \sqrt{3} {x}^{2} - 3 x - 3 \sqrt{3}\right) + \left(0\right)$

#### Explanation:

$\left({x}^{4} - 6 {x}^{2} + 9\right) \div \left(x - \sqrt{3}\right)$

Using synthetic division :

We have , $p \left(x\right) = \left({x}^{4} + 0 {x}^{3} - 6 {x}^{2} + 0 x + 9\right) \mathmr{and} \text{divisor : } x = \sqrt{3}$

We take ,coefficients of $p \left(x\right) \to 1 , 0 , - 6 , 0 , 9$

$\sqrt{3} |$ $1 \textcolor{w h i t e}{\ldots \ldots .} 0 \textcolor{w h i t e}{\ldots \ldots} - 6 \textcolor{w h i t e}{\ldots \ldots . .} 0 \textcolor{w h i t e}{\ldots \ldots . .} 9$
$\underline{\textcolor{w h i t e}{\ldots .}} |$ ul(0color(white)( ....)sqrt3color(white)(........)3color(white)(...)-3sqrt3color(white)(...)-9
color(white)(......)1color(white)(......)sqrt3color(white)(....)-3color(white)(..)-3sqrt3color(white)(....)color(violet)(ul|0|
We can see that , quotient polynomial :

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

Hence ,

$\left({x}^{4} - 6 {x}^{2} + 9\right) = \left(x - \sqrt{3}\right) \left({x}^{3} + \sqrt{3} {x}^{2} - 3 x - 3 \sqrt{3}\right) + \left(0\right)$