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

##### 1 Answer

By showing that the division gives no remainder, i.e.

The basic setup of **synthetic division** is:

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

#" "+ " "ul(" "" "" "" "" "" "" "" "" "" "" "" "" ")#

where the *coefficients* correspond to the function *quadratic* back.

(Note that the root you use in your divisor would be the root you acquire from

The general steps are:

- Bring down the first coefficient.
- Multiply the result beneath the horizontal line by the root and store in the next column above the horizontal line.
- Add that column.
- 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")" "" ")#

#" "" "" "" "1" "(2+sqrt2)" "2^"3/2"" "" "" "0#

Your resultant quadratic then should be:

#= color(blue)(x^2 + (2 + sqrt2)x + 2^"3/2")#

So, your solution would be:

#(x^3 + 2x^2 - 2x - 4)/(x - sqrt2) = stackrel(p(x)"/"q(x))overbrace(x^2 + (2 + sqrt2)x + 2^"3/2") + stackrel(r(x))overbrace(0/(x - sqrt2))#

And you can check to see that it properly expands to give the original cubic. Furthermore, we now see that

*CHALLENGE: Can you show that* *is also a root?*