Question

Factor the polynomial: 32a^2b^2+16ab^2+2b^2 MEDAL WILL BE GIVEN TO THE BEST EXPLANATION!

Answer 1

if u want to learn, at least try simplifying your equation a bit. by dividing everything by 2.

Answer 2

or expression for that matter.

Answer 3

wouldnt i divide by 4??

Answer 4

Use grouping, but factor out the obvious common factors first:
\( 32a^2b^2+16ab^2+2b^2 \\
=2b^2(16a^2+8a+1)
\)
Now factorize what's inside the parentheses, find m and n such that m+n=8, m*n=16*1=16
Proceed to grouping as in the other problem.

Answer 5

last term is 2 so the gcf is 2

Answer 6

@mathmate you can also factor out the a though?

Answer 7

Sorry, but I don't see an \(a\) in the last term \(2b^2\), so \(a\) is not a common factor.

Answer 8

@mathmate can you go over the "factorize inside the parentheses" part, you sorta lost me there

Answer 9

Use grouping.
Let ax^2+bx+c be the expression to be factorized.
We look for integers m, n such that m+n=b, and m*n=a*c.
For the given problem, a=16, b=8, c=1.
So we look for m,n such that m+n=8, m*n=16*1=16

Can you find m, n satisfying these conditions?

Answer 10

@ineedyouubiebs guess you're not online.
I will explain the other steps, so you can continue when you're online.

Here, we see that m=n=4 to give mn=16, m+n=8
So we group the expression as follows:
16x^2+4x + 4x + 1
=4x(4x+1) + (4x+1)
take out the common factor (4x + 1) and you'll get
=(4x+1)(4x+1)
=(4x+1)^2

In this particular case, there is a shorter method:
If both the x^2 term and the constant term are perfect squares, as in
(4x)^2=16x^2, and
1^2 = 1
We double the product (1*4x) to get 8x and compare with the middle term. If they equal, then the expression is a perfect square, equal to (4x+1)^2.

Answer 11

http://library.thinkquest.org/20991/alg2/polyf.html