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

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.
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")" "" ")
$\text{ "" "" "" "1" "(2+sqrt2)" "2^"3/2"" "" "" } 0$

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