# Dividing polynomial fractions with varying term quantities I'm working through an old algebra book as a refresher and I've come across what should be a simple polynomial division. The exercise prompts the reader to perform the following operation: (a^2-9)/(a^2+3a) -: (a-3)/(4)

jatericrimson8b 2022-09-12 Answered
Dividing polynomial fractions with varying term quantities
I'm working through an old algebra book as a refresher and I've come across what should be a simple polynomial division. The exercise prompts the reader to perform the following operation:
$\frac{{a}^{2}-9}{{a}^{2}+3a}÷\frac{a-3}{4}$
I started off by inverting the $÷$ sign by instead multiplying by the reciprocal which results in:
$\frac{{a}^{2}-9}{{a}^{2}+3a}\cdot \frac{4}{a-3}$
I spent nearly 2 hours at this point trying everything my mind could conjure in terms of factoring, simplifying, and multiplying, but none of my attempts ever arrived at the listed answer:
$\frac{4}{a}$
If someone could help me through the steps required to solve this, you'll have taught a man to fish.
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## Answers (2)

Conner Singleton
Answered 2022-09-13 Author has 13 answers
Because
$\frac{{a}^{2}-9}{{a}^{2}+3a}\cdot \frac{4}{a-3}=\frac{4\left(a-3\right)\left(a+3\right)}{a\left(a+3\right)\left(a-3\right)}=\frac{4}{a}$

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restgarnut
Answered 2022-09-14 Author has 1 answers
note that
${a}^{2}-9=\left(a-3\right)\left(a+3\right)$
${a}^{2}+3a=a\left(a+3\right)$

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