# How do you use the remainder theorem to find the remainder for each division (x^4-6x^2+8)div(x-sqrt2)?

Dec 1, 2017

Remainder is $0$

#### Explanation:

let $f \left(x\right) = {x}^{4} - 6 {x}^{2} + 8$

and $g \left(x\right) = x - \sqrt{2}$

Then:$f \left(x\right) = g \left(x\right) q + r$

Where $q$ and $r$ are the quotient and remainder respectively.
This is The Remainder Theorem

It can be seen from this, that if we can make $g \left(x\right) = 0$, then we can find the remainder $r$.

$\therefore$

${x}^{4} - 6 {x}^{2} + 8 = q \left(x - \sqrt{2}\right) + r$

Let $x = \sqrt{2}$

${\left(\sqrt{2}\right)}^{4} - 6 {\left(\sqrt{2}\right)}^{2} + 8 = q \left(\sqrt{2} - \sqrt{2}\right) + r$

$4 - 12 + 8 = q \left(0\right) + r \implies r = 0$

So the remainder is $0$

There is only ever 1 remainder.