How do you divide $( 2x^4 - 5x^3 - 8x^2+17x+1 )/(x^2 - 2 )$ ?

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Answer 1 · 2017-06-14T00:35:32

$(2 x^4 - 5 x^3 - 8 x^2 + 17 x + 1 )/(x^2 - 2)= (2 x^2 - 5 x - 4) + (7 x - 7 )/(x^2-2)$

The first thing we should do when trying to divide two rational functions is to see if the denominator is a zero of the numerator.

Let $f(x)=2x^4-5x^3-8x^2+17x+1$

$f(sqrt2)=-7+7sqrt2$

We know that any rational function can be written as: $f(x)=p(x)q(x)+r(x)$ . Thus we can say:

$2 x^4 - 5 x^3 - 8 x^2 + 17 x + 1 =(ax^2+bx+c)(x^2-2)+7x-7$

This is because when $(x^2-2)=0$ , the remainder was $(7x-7)$ . So $r(x)=7x-7$ . We already know $q(x)$ and we need to work out $p(x)$ .

$2x^4-5x^3-8x^2+10x+8=(ax^2+bx+c)(x^2-2)$

$ax^4=2x^4rArra=2$

$bx^3=-5x^3rArrb=-5$

$-2c=8rArrc=-4$

$2x^4-5x^3-8x^2+10x+8=(2x^2-5x-4)(x^2-2)$

$2 x^4 - 5 x^3 - 8 x^2 + 17 x + 1=(2x^2-5x-4)(x^2-2) +7x-7$

$(2 x^4 - 5 x^3 - 8 x^2 + 17 x + 1 )/(x^2 - 2)= (2 x^2 - 5 x - 4) + (7 x - 7 )/(x^2-2)$

How do you divide ( 2x^4 - 5x^3 - 8x^2+17x+1 )/(x^2 - 2 )( 2x^4 - 5x^3 - 8x^2+17x+1 )/(x^2 - 2 )?