# How do you use synthetic division to show that x=sqrt2 is a zero of x^3+2x^2-2x-4=0?

##### 1 Answer
Jan 24, 2017

By showing that the division gives no remainder, i.e. $r \left(x\right) = 0$.

The basic setup of synthetic division is:

ul(sqrt2" ") "|"" "" "1" "" "2" "" "-2" "" "" "-4
" "+ " "ul(" "" "" "" "" "" "" "" "" "" "" "" "" ")

where the coefficients correspond to the function ${x}^{3} + 2 {x}^{2} - 2 x - 4$. Since we are dividing by a linear factor, $x - \sqrt{2}$, we should get a quadratic back.

(Note that the root you use in your divisor would be the root you acquire from $x - r = 0$.)

The general steps are:

1. Bring down the first coefficient.
2. Multiply the result beneath the horizontal line by the root and store in the next column above the horizontal line.
3. Add that column.
4. Repeat 2-3 until you reach the last column and have evaluated the final addition.

You should then get:

ul(sqrt2" ") "|"" "" "1" "" "2" "" "-2" "" "" "-4
" "+ " "ul(" "" "" "" "sqrt2" "(2^"3/2" + 2)" "2^("1/2" + "3/2")" "" ")
$\text{ "" "" "" "1" "(2+sqrt2)" "2^"3/2"" "" "" } 0$

Your resultant quadratic then should be:

$= \textcolor{b l u e}{{x}^{2} + \left(2 + \sqrt{2}\right) x + {2}^{\text{3/2}}}$

So, your solution would be:

$\frac{{x}^{3} + 2 {x}^{2} - 2 x - 4}{x - \sqrt{2}} = \stackrel{p \left(x\right) \text{/"q(x))overbrace(x^2 + (2 + sqrt2)x + 2^"3/2}}{+} \stackrel{r \left(x\right)}{\overbrace{\frac{0}{x - \sqrt{2}}}}$

And you can check to see that it properly expands to give the original cubic. Furthermore, we now see that $r \left(x\right) = 0$, so it is evident that $x = \sqrt{2}$ is a root of ${x}^{3} + 2 {x}^{2} - 2 x - 4$.

CHALLENGE: Can you show that $- \sqrt{2}$ is also a root?