How do you divide $\frac{8 x^{2} - 10 x - 7}{4 x + 1}$ $(8x^2-10x-7)/(4x+1)$ ?

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How do you divide $\frac{8 x^{2} - 10 x - 7}{4 x + 1}$ $(8x^2-10x-7)/(4x+1)$ ?

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Answer 1 · 2016-07-03T19:02:52

$\frac{8 x^{2} - 10 x - 7}{4 x + 1} = 2 x - 3 - \frac{4}{4 x + 1}$ $(8x^2 - 10x - 7)/(4x+1) = 2x - 3 - 4/(4x+1)$

Explanation:

This is very hard to format on here so I'm going to assume you sort of know how to do polynomial long division and just walk you through the steps. I can hand write it and scan it in if you need more help visualising what I'm on about.

Dividing 4x into first term gives 2x.

Multiply $4 x + 1$ $4x+1$ by the $2 x$ $2x$ and subtract, giving:

$8 x^{2} - 10 x - 7$ $8x^2 - 10x - 7$
$- [8 x^{2} + 2 x]$ $- [8x^2 + 2x]$

$= - 12 x - 7$ $= -12x - 7$

Dividing in again gives -3. Multiply by -3 and subtract:

$- 12 x - 7$ $-12x-7$
$- [- 12 x - 3]$ $- [-12x - 3]$

$= - 4$ $= -4$ which is our remainder.

$\frac{8 x^{2} - 10 x - 7}{4 x + 1} = 2 x - 3 - \frac{4}{4 x + 1}$ $(8x^2 - 10x - 7)/(4x+1) = 2x - 3 - 4/(4x+1)$

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How do you divide $\frac{8 x^{2} - 10 x - 7}{4 x + 1}$ $(8x^2-10x-7)/(4x+1)$ ?

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