Factoring with Repeated Use of Difference of Squares Formula

I need help with this problem: "u^4w^2 - 16w^2

This is the answer I came up with, but it does not look right: u^4(w+8)(w-8)

Thanks

ondverg said:

I need help with this problem: "u^4w^2 - 16w^2

This is the answer I came up with, but it does not look right: u^4(w+8)(w-8)

Thanks

Click to expand...

Did you check your answer by multiplying the factors together? If you do NOT get the expression you started with, you can be sure your factorization is not correct.

Did you notice that both terms have a factor of w[sup:187uhtf4]2[/sup:187uhtf4]? I'd start by using the Distributive Property to remove that common factor. Then, look at the expression you get to see if you can apply the "difference of two squares" formula.

If you're still having trouble, please repost showing ALL of your work so that we can see where you might be making an error.

Answer: u^4(w+4w)(w-4w)
Check: = u^4 w^2 - 16w^2

Is this right?

Thanks

Notice the perfect squares.

\(\displaystyle (u^{2}w)^{2}-(4w)^{2}\)

Now, just factor using the difference of two squares factorization.

Thank you so much! I understand your answer perfectly and believe that I can do the problems now. What had me confused is the way the samples are done in my booklet. It give the answers to a few problems, but doesn't show how it's done. Here's an example:

Problem:16x^2 - x^2y^4
Answer: x^2(4+y^2) (2+y) (2-y)

I was attempting to get my answer to look similar to the above format, but it's appears more confusing whereas your answer to the problem I had to work is clear.

Thanks again!

ondverg said:

I need help with this problem: "u^4w^2 - 16w^2

This is the answer I came up with, but it does not look right: u^4(w+8)(w-8)

Thanks

Click to expand...

You could also look at the following way:

\(\displaystyle u^4w^2 - 16w^2 = w^2 [ u^4 - 16] = w^2[ (u^2)^2 - (2^2)^2] = w^2[(u^2 + 2^2)(u^2 - 2^2)] = w^2 (u^2 + 4)(u + 2)(u - 2)\)