Can you help me factor a^3+2a^2-ab^2-2b^2?

Question

The book tells me the answer is (a+b)(a-b)(a+2), but I cannot figure out how to get that answer... Every method I try gives me something different.. What am I doing wrong??

First, I tried grouping "a^3+2a^2"-"ab^2-2b^2" which turned into a^2(a+2a)-b(ab-2b) and got (a^2-b)(a+2a)(ab-2b). This is wrong..

So I realized... I could group a^3+ "2a^2-ab^2-2b^2" ...when I move that first 2 to the second 2 I get a^2-ab^2-4b^2... But I get stuck here... What am I doing wrong?!

It might be more clear to read if you write it down, "a^3" is "a to the power 3"

Thank you for your time!

Oh, you're right, Lola... That was a typo, I have on my paper (a+2), not (a+2a)

Samuel D, how did you get those exponant numbers?!

Thanks Brett, looks like this expression was set up to make an example of people like me!

Nevermind Lola, that wasn't a typo... I did this expression so many times, I had it right on one, and wrong on another, and I copied the wrong one... My other mistakes were continually made though. Hmm

Answer 1

You were close with your first attempt. It turns into a^2(a+2)-b^2(a+2), which becomes (a^2-b^2)(a+2), which becomes the answer.

Your first mistake was grouping it wrong. It's not
"a^3+2a^2"-"ab^2-2b^2"
It's
"a^3+2a^2"-"ab^2+2b^2". Putting the quotes around the last two terms means the '-' in front of the quotes applies to each term inside (-ab^2 and to +2b^2). When the negative outside the quotes hits the +2b^2, it becomes the -2b^2 in the question.

Your 2nd mistake: you only pulled one b out of "ab^2-2b^2". They have a b^2 in common, so pull out a b^2, making b^2(a-2). If you had done that, you might have noticed that the (a-2) looks a lot like the (a+2a) result of the first two terms.

Which is your 3rd and final mistake. It does not turn into
a^2(a+2a)-b(ab-2b)
It turns into
a^2(a+2)-b^2(a+2).
When you pulled the a^2 out of the +2a^2 term, you left an 'a' behind.

I hope this makes sense.

Answer 2

The first thing you tried was the right idea, except that when you factored out the a^2, that would leave you with (a + 2), NOT (a + 2a). Notice that when you take the a^2 out of 2a^2, you're just left with 2. Also, the last 2 terms have a b^2 in common, not just b. So you can pull out the b^2, but be careful with the minus sign. Notice that in what you tried, you pulled out a "negative b", not just "b", so your 2nd term in parentheses should have been +2b, not -2b.

So, applying all of that, let's pull a^2 out of the first 2 terms and a -b^2 out of the last 2 terms:
a^3+2a^2-ab^2-2b^2
a^2(a+2) - b^2(a+2)

Now you can probably see where this is going - both terms now have (a+2) in common, so we can pull that out:
(a^2 - b^2)(a+2)

Now reverse FOIL the (a^2 - b^2), and you get:
(a+b)(a-b)(a+2)

Answer 3

Not all can be done this way, but I think you can factor by grouping:

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

Edit: To keep from making a mistake with the negative, I ALWAYS put + between my groups and keep the sign of the third term with the third term. Then, if necessary, I factor out the negative.

Factor out the gcf in both groups:

a^2 (a+2) - b^2(a+2)

Factor out (a+2):

(a+2)(a^2 - b^2)

Factor the second parentheses:

(a+2)(a+b)(a-b)

Hope this helps.

Answer 4

a^3+2a^2 - ab^2 - 2b^2 groups into
a^2(a + 2) - b^2(a + 2) =
(a + 2)(a^2 - b^2) = (difference of squares now)
(a + 2)(a - b)(a + b)

Answer 5

a^2 (a+2) - b^2 (a+2)

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

Answer 6

a³ + 2a² - ab² - 2ab²

a²(a + 2) - b²(a + 2)

(a² - b²)(a + 2)

(a - b)(a + b)(a + 2)

- - - - - - - - - -s-