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

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

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Answer 1 · 2017-06-14T00:35:32

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

Explanation:

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

$f (\sqrt{2}) = - 7 + 7 \sqrt{2}$ $f(sqrt2)=-7+7sqrt2$

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

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

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

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

$a x^{4} = 2 x^{4} \Rightarrow a = 2$ $ax^4=2x^4rArra=2$

$b x^{3} = - 5 x^{3} \Rightarrow b = - 5$ $bx^3=-5x^3rArrb=-5$

$- 2 c = 8 \Rightarrow c = - 4$ $-2c=8rArrc=-4$

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

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

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

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

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How do you divide 2x4−5x3−8x2+17x+1x2−2( 2x^4 - 5x^3 - 8x^2+17x+1 )/(x^2 - 2 )?

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