Expert: Richard J. Raridon - 11/7/2009

Question
QUESTION: Good Afternoon,

I'm an older working mom planning on returning to school next Spring, and I am teachng myself algebra.

Here is the problem:

a^2+ab-b^2 -(a^3-2b^3/a-2b)

Since I cannot factor a^+ab-b^2 further, I multiply it by
the denominator(a-2b)= a^3-2a^2b + a2b-2ab^2 - ab^2+2b - (a^3-2b^3/a-2b)

I then combine like terms thusly:

a^3      -2a^2b     -2ab^2     +2b^3
-a^3      + a^2b     - ab^2     -2b^3

= -a^2b -3ab^2

My answer does not match the result in my textbook.

Please enlighten me with your expertise.  Thank you in advance.

ANSWER: The 2b^3's are both positive, plus you lost the denominator.

---------- FOLLOW-UP ----------

QUESTION: Thanks for your quick reply.

However, in working my problem again, I don't see how the 2b^3's are both positive. Also, even if I were to make them both positive - the answer still does not match the book answer.

Additionally, 3ab^2 is positive in my textbook.

As for the denominator, it was an oversight not to include it - thanks for pointing it out.

Here is the answer that my textbook provides:
a^2b +3ab^2 -4b^3/a-2b (note: the sign of the fraction is negative, I just don't know how to show it on here.)

My gut tells me I need to change my signs around, but I don't know why or which ones.


ANSWER: It's not good practice to write the answer that way.  It should be written
(a^2b+3ab^2-4b^3)/(a-2b)   
If you wrote the problem correctly originally, then that answer is not correct.  It should be multiplied by -1 which is what you had for the first two terms.
The reason the 2b^3's are both positive is because, in both cases, you're multiplying two negatives.

I have taught physics and I know the answers in the back are not always correct.  Usually they get graduate students to work the problems or proof them.

---------- FOLLOW-UP ----------

QUESTION: Well now I see that I was in error, sir.

I presented the problem incorrectly. The corrected problem is a^2+ab-b^2 -(a^3-2b^3/a-2b)

Re: your statement about multiplying the 2 terms in the numerator by -1; well, I hadn't done that at all and it caused me to look closely at the original problem.  The sign of the fraction itself is negative!(They say the devil is in the details - but I digress.)

At any rate, I was able to arrive at the answer in my textbook, by reversing the signs in the numerator AND the sign of the fraction (thus preserving the value of said fraction) before performing the multiplication of the trinomial by the denominator.

My question is this:  Why change the signs to begin with? Is it because a negative fraction is considered unwieldy or because the numerator is not in descending order of value?  Is there some rule I'm not aware of that caused my blindspot of neglecting to reverse the signs?

Again thank you for your assistance.  I look forward to your reply.

Answer
I'm older than you but you don't need to call me sir.  Your corrected problem looks exactly like your original problem.  Am I missing something?  There's no rule that you have to change signs to begin with and the terms in the numerator can be in any order.  You can't just reverse the signs in the numerator and the sign of the fraction without reversing the sign in the denominator.  I stand by my statement that the answer in the book is wrong.