Factoring Help Needed :(

[xxxegoxxx]

1. (x^2+5x)^2 - 36
2. 2(2x^2-x)^2 - 3(2x^2-x) - 9
3. 6(a+b)^2+17(a+b)+5
4. (x^2+3x)^-2(x^2+3x)-8

Show the steps please, I don't know what to do with the random ass number at the end

Edit: I have figured out the solutions, you can let the polynomial represent x in the bracket, and then plug it in at the end.

 

[Weasel2013]

1. (x^2+5x)^2 - 36
2. 2(2x^2-x)^2 - 3(2x^2-x) - 9
3. 6(a+b)^2+17(a+b)+5
4. (x^2+3x)^-2(x^2+3x)-8

Show the steps please, I don't know what to do with the random ass number at the end

It's a difference of two squares. However, I got discouraged after #1. the binomial ends up to be X(x+5)^2, which cannot be factored..so i'm guessing it's prime?

 

[xxxegoxxx]

Yes, it is difference of squares, but there's a random number at the end that throws me off...

Edit: I don't know what you mean by prime, but it is possible to factor all of these, because there's like half a page of similar problems.

 

[me_person666]

If you want some VERY good help, my old math teacher has some really good math videos.

http://www.youtube.com/user/minkusbc#p/u

It helped me so much for my Principles math 10 provincial exam.

 

[Vanmaster5]

1. (x^2+5x)^2 - 36 = x^4 + 25x^2 - 36
2. 2(2x^2-x)^2 - 3(2x^2-x) - 9 = 2(4x^4-x) -6x^2+3x = 8x^6 - 6x^2 + x
3. 6(a+b)^2+17(a+b)+5 = 6a^2 + 6b^2 + 17a + 17b
4. (x^2+3x)^-2(x^2+3x)-8 = 1/x^4 + 9x^2 (-8x^2 - 24x), then FOIL it, i think u may have wrote the problem wrong bc 1-3 are quite easy then number 4 is going to be a screwed up huge number.