When you factor, you first factor anything out that you can, then cancel the common factors.

ex.

35v – 14

_______

49v

7(5v – 2)

____

49v

Factor out the GCF of the numerator, which means you can pull out a 7 from the top and bottom.

If you do that, you can cancel parts out:

5v – 2

______

7v

See, the 49/7 became 7v in the denominator, and the 5v because it cancelled with the numerator's 7.

That should help you to figure it out, without giving you the exact answer.

When you add in an exponent, you do the same thing, but factor out the exponents too.

For ex.

7f^2+f

______

7f+1

7f^2+f can be broken down into

f times (use parenthesis instead of times around the next part,) 7f+1 over (use / instead of over) 7f+1. So, now you can cancel out the factored expression (7f+1) with the denominator, and you're left with f!

With yours, I'll help you with the first part...the z part.

See if you can figure out the second part of the question you're supposed to answer - the factorization of the y part (part 2). You can factor out a z, so you can cancel both the top and bottom z's, and no, I'm not sleeping...zzzz. ;)

z(z^4)y6/zy^4

The z's cancel out and you end up with z^4(y^2) as the first answer. It's y^2nd power (still part of 1) because you subtract the denominator's exponent from the numerator's exponent (using the like terms...the y's). So...y^6-y^4=y^2. Now, all you have to do is figure out the 2nd part of the question. Be careful, your teacher either goofed or she is trying to snag you, or you wrote out the question incorrectly. You can't factor out a y^5 unless there is a y^5th there...unless, of course, you meant: z^7y^5/zy^4 was the original problem she gave you?