# Pesky Polynomial Multiplication Problem

• Jul 13th 2010, 11:57 AM
Ingersoll
Pesky Polynomial Multiplication Problem
3a(a -2b)^2. It seems to me that you'd just multiply the numbers inside the parentheses by themselves.

That would give you 3a(a^2+4b); the answer to that is 3a^3 +12ab^2. But that's not the right answer.

The right answer is 3a^3 -12a^2b +12ab^2.

So what am I missing?
• Jul 13th 2010, 12:21 PM
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Quote:

Originally Posted by Ingersoll
3a(a -2b)^2. It seems to me that you'd just multiply the numbers inside the parentheses by themselves.

That would give you 3a(a^2+4b); the answer to that is 3a^3 +12ab^2. But that's not the right answer.

The right answer is 3a^3 -12a^2b +12ab^2.

So what am I missing?

(a -2b)^2 = (a - 2b)(a - 2b) = a^2 - 4ab + 4b^2

Multiply that by 3a
• Jul 13th 2010, 12:31 PM
HallsofIvy
Quote:

Originally Posted by Ingersoll
3a(a -2b)^2. It seems to me that you'd just multiply the numbers inside the parentheses by themselves.

That would give you 3a(a^2+4b); the answer to that is 3a^3 +12ab^2. But that's not the right answer.

The right answer is 3a^3 -12a^2b +12ab^2.

So what am I missing?

What you are missing is that \$\displaystyle (a- 2b)^2\$ is NOT just \$\displaystyle a^2- (2b)^2\$. \$\displaystyle (a- 2b)^2\$ means (a- 2b)(a- 2b)= a(a- 2b)- 2b(a- 2b) (the "distributive law") which is equal to a(a)+ a(-2b)- 2b(a)- 2b(-2b) (the distributive law used again- twice) and that is equal to \$\displaystyle a^2- 4ab+ 4b^2\$ as incomplete said.
• Jul 13th 2010, 12:39 PM
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Quote:

Originally Posted by HallsofIvy
...as incomplete said.

(Rofl) (Rofl) I type quickly on IM programs and produce some funny typos, but this one really made me laugh!
• Jul 13th 2010, 12:59 PM
wonderboy1953
For Ingersoll:

It's always best to break up a problem into parts whenever you can.

What I would have done was to temporarily cover up the 3a, then expand the parentheses by squaring it, then uncover the 3a to multiply the result from squaring the parentheses and you would have gotten your answer. This method can be adapted towards different algebra problems.
• Jul 13th 2010, 01:24 PM
Ingersoll
Thanks. Really, I'm not that dumb. Maybe trying to burn through 3 straight days of homework to get ahead isn't such a good idea. Wow, my brain is foggy. But thanks again. :)