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

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Answer 1 · 2017-12-01T17:27:38

Remainder is $0$

let $f(x)=x^4-6x^2+8$

and $g(x)=x-sqrt(2)$

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

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 $g(x)=0$ , then we can find the remainder $r$ .

$:.$

$x^4-6x^2+8=q(x-sqrt(2))+r$

Let $x=sqrt(2)$

$(sqrt(2))^4-6(sqrt(2))^2+8=q(sqrt(2)-sqrt(2))+r$

$4-12+8=q(0)+r=>r=0$

So the remainder is $0$

There is only ever 1 remainder.

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