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This is a really low level question. I'm trying to simplify $$f(x) = \frac{36x^{-2} -3x^{-1} -18}{12x^{-2} -25x^{-1} +12}$$

After factoring, removing negative exponents, and flipping the second fraction I get $$f(x) = \frac{1}{3(3x+2)(4x-3)} \frac{(3x-4)(4x-3)}{1}$$

Then $(4x-3)$ cancels leaving

$$f(x) = -\frac{3x-4}{3(3x+2)}$$

as my final answer. However the book I have says the correct answer is $$f(x) = -\frac{3(2x+3)}{4x-3}$$

I've checked my work many times and I don't know how they get this answer. Could someone please help me solve this?

Wanatabe
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2 Answers2

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I suggest getting rid of the negative exponents first: $$\eqalign{\frac{36x^{-2}-3x^{-1}-18}{12x^{-2}-25x^{-1}+12} &=-\frac{18x^2+3x-36}{12x^2-25x+12}\cr &=-\frac{3(3x-4)(2x+3)}{(3x-4)(4x-3)}\cr &=-\frac{3(2x+3)}{4x-3}\ .\cr}$$

David
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Your error came when you factored $36x^{-2} -3x^{-1} -18$ as $(36x^2 -3x^1 -18)^{-1}$ which is not the same as $36x^{-2} -3x^{- 1} -18$, the correct way to solve comes when you realize that $36x^{-2}$ is the same as $36/x^2$, seeing as this is the case you multiply the fraction by $-x^2$ and get −$18x^2+3x−36/12x^2−25x+12$ and David's solution takes it from there.

Davide Giraudo
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Matthew B
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