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In order to do long division on this problem, do I have to first do Foil? $$\frac{6x^2-3x}{(x-2)(x+4)}$$ How should I go about this long division?
If I do FOIL first, do I end up with a problem like this? $$(x^2+4x-8)\overline { )6x^2-3x}$$How should I do the division on it?

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    $\begingroup$ Yea, FOIL first, then divide. $\endgroup$ – Don Thousand Sep 15 '19 at 22:31
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The remainder has degree at most $1$, so write it in the form $ax+b$. Then $$ 6x^2-3x=(x-2)(x+4)Q(x)+ax+b $$ Evaluating at $2$ yields $18=2a+b$; evaluating at $-4$ yields $108=-4a+b$.

Therefore $90=-6a$ and $a=-15$; finally $b=18+30=48$.

The remainder is $-15x+48$.

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$$\frac{6x^2-3x}{(x-2)(x+4)}$$

Expand (FOIL):

$$\frac{6x^2-3x}{x^2+2x-8}$$

When dividing, you get a quotient of $6$ and a remainder of $-15x+48$. Use the polynomial division formula to get:

$$6+\frac{-15x+48}{x^2+2x-8}$$

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