# Long Division in Fractions

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?

• Yea, FOIL first, then divide. – Don Thousand Sep 15 '19 at 22:31

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 $$-15x+48$$.

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

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:

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