i am terrible at factoring and i would like some much needed help with these!

a) 36(2x - y)^2 - 25(u - 2y)^2
b) x^2 - y^2 + z^2 - 2xz
c) x^2 + 2 + 1/x^2
d) p^2 - 2p + 1 - y^2 - 2yz - z^2

thanks!!!

2. Originally Posted by checkmarks
c) x^2 + 2 + 1/x^2
d) p^2 - 2p + 1 - y^2 - 2yz - z^2
c) $x^2 + 2 + \frac{1}{x^2}$

If this were $a^2 + 2ab + b^2$ you would have no trouble factoring it (I presume).

Let $a = x$ and $b = \frac{1}{x}$

$a^2 + 2ab + b^2 = (a + b)^2 = \left ( x + \frac{1}{x} \right ) ^2$

This is a useful little trick to remember. You'll see this kind of factoring every now and again.

d)
$p^2 - 2p + 1 - y^2 - 2yz - z^2$

$(p^2 - 2p + 1) - (y^2 + 2yz + z^2)$

$(p - 1)^2 - (y + z)^2$

Nothing more you can do here.

-Dan

3. Originally Posted by checkmarks
a) 36(2x - y)^2 - 25(u - 2y)^2
b) x^2 - y^2 + z^2 - 2xz
The first one can be solved using the identity $a^2-b^2=(a+b)(a-b)$

The second one it's similar, look for a perfect square, then factorice.

Originally Posted by topsquark
d)
$p^2 - 2p + 1 - y^2 - 2yz - z^2$

$(p^2 - 2p + 1) - (y^2 + 2yz + z^2)$

$(p - 1)^2 - (y + z)^2$

Nothing more you can do here.
You sure?

4. Originally Posted by Krizalid
The first one can be solved using the identity $a^2-b^2=(a+b)(a-b)$

The second one it's similar, look for a perfect square, then factorice.
i figure that but i dont know what to do with the 36 and 25

5. Originally Posted by checkmarks
a) 36(2x - y)^2 - 25(u - 2y)^2
$36(2x-y)^2-25(u-2y)^2=[6(2x-y)]^2-[5(u-2y)]^2$

Now apply $a^2-b^2=(a+b)(a-b)$

6. Originally Posted by Krizalid
The first one can be solved using the identity $a^2-b^2=(a+b)(a-b)$

The second one it's similar, look for a perfect square, then factorice.

You sure?

So just how many times have you had to correct me today, Krizalid?

-Dan

7. I hope you don't get mad.

There're just details, it's nothing to worry about

8. Originally Posted by Krizalid
I hope you don't get mad.

There're just details, it's nothing to worry about
I only get mad at myself. I don't mind making the occasional mistake, because everyone does at some point and because there are skilled workers here who will catch them.

What I get frustrated about is when I keep making mistakes over and over and over... That means I'm simply working too fast and that doesn't really help the people who have asked for it.

(Not to mention I have nearly as god-like an ego as ThePerfectHacker. Don't tell him that, though, because he might feel that there isn't enough Hilbert space here for the two of us! )

-Dan