# Algebra II Quotients of Rational Expressions

1. Mar 21, 2012

### velox_xox

1. The problem statement, all variables and given/known data
Simplify. (p4 - q4)/(p + q)2 ÷ 1/(p2 + q2)

Answer: (p - q)/(p + q)

2. Relevant equations
--

3. The attempt at a solution
Transformed it to a multiplication problem.
(p4 - q4)/(p + q)2 X (p2 + q2)/1
Difference of the squares in the numerator of the first expression:
(p2 + q2)(p2 - q2)/(p + q)2 X (p2 + q2)/1
Difference of the squares (once more) in the numerator of the first expression:
(p2 + q2)(p + q)(p - q)/(p + q)2 X (p2 + q2)/1
Simplified the (p +q) from the numerator and denominator of the first equation:
(p2 + q2)(p -q)/(p + q) X (p2 + q2)/1
Simplified the (p2 + q2) from the numerators of both expressions:
(p - q)/(p + q)

My question is with the part. Is that proper form? A.k.a. Did I not bend the rules of algebra to get the correct answer? And if so, can someone give example of when simplifying both numerators is okay or not okay?

Thank you!

Last edited: Mar 21, 2012
2. Mar 21, 2012

### Dick

Now if (p^2+q^2) is in the numerators of both expressions how can you just cancel it out? Shouldn't your answer have (p^2+q^2)^2 in it?

3. Mar 21, 2012

### velox_xox

That's what I thought, but the textbook's answer is (p - q)/(p +q), so either it's incorrect; Or somewhere I solved it wrong in the first place, and now I'm just bending the rules. That's why I'm asking. :)

4. Mar 21, 2012

### Dick

Well, I get (p-q)*(p^2+q^2)^2/(p+q). I don't think you are doing anything wrong except for the bad cancellation to match the books answer. Either the books answer is wrong or the problem is misstated.

5. Mar 21, 2012

### velox_xox

Nope, I've checked it like seven times now to be sure (both the problem and the answer). The only thing that is different in my writing of the problem from the actual textbook's version is that it is actually a fraction within a fraction.

(p4 - q4)/(p + q)2 <<This part being numerator
(p2 + q2) <<This part being denominator

That shouldn't make a difference, though. Right?

...Anyone else wanna give it a go?

6. Mar 22, 2012

### scurty

Let's try writing them as fractions. From the post I quoted I get the impression the fraction is: $\displaystyle\frac{\frac{p^4-q^4}{(p+q)^2}}{p^2+q^2}$. If this is the case, you gave us the wrong problem to work with in your original post!

$\displaystyle\frac{\frac{p^4-q^4}{(p+q)^2}}{p^2+q^2} = \frac{p^4-q^4}{(p+q)^2} \cdot \frac{1}{p^2+q^2} \neq \frac{p^4-q^4}{(p+q)^2} \div \frac{1}{p^2+q^2}$

7. Mar 22, 2012

### velox_xox

*facepalm* Oh, that is so obvious! Like how 8 and 1/8 aren't the same.

I was able to get the correct answer with that. Also, for future reference, how am I supposed to write a fraction in the numerator of a fraction?? Like what you did? (The reason I didn't write that it that way in the first place is because I didn't know how to do it.)

Thanks scurty for your insight, and sorry Dick for the mistake!

8. Mar 22, 2012

### Dick

No problem. Just use an extra set of parentheses. Like ((p^4-q^4)/(p+q)^2))/(p^2+q^2). Or learn to Latex it. That's always nice.

9. Apr 20, 2012

### velox_xox

Got it. I'll be sure to do that in the future. As for Latex, I have no idea how to do that. Is there a section on how to learn on the forum? :D

And a very much belated but definitely deserved thanks to scurty and Dick. Thank you!

10. Apr 21, 2012