1.) 36x+18xy-9y-18

2.) 4-36x^2

3.) 42x^2-24

4.) 12x^2-5xy-3y^2

5.) -14r^2-11rs+15s^2

6.) x^4-x^2-30

I'm really struggling with this so any help would be greatly appreciated!

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- Sep 17th 2008, 03:20 PMepetrikFactoring! YUCK!
1.) 36x+18xy-9y-18

2.) 4-36x^2

3.) 42x^2-24

4.) 12x^2-5xy-3y^2

5.) -14r^2-11rs+15s^2

6.) x^4-x^2-30

I'm really struggling with this so any help would be greatly appreciated! - Sep 17th 2008, 03:29 PMJhevon
factor by grouping. there is a common 18x in the first two terms, and a common -9 in the last two

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2.) 4-36x^2

now apply the formula for the difference of two squares factorization

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3.) 42x^2-24

now, what do you aee?

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4.) 12x^2-5xy-3y^2

here's a tip: forget the y for a while and treat it as $\displaystyle 12x^2 - 5x - 3$ and factor that. when finished, attach a y to the constant terms, and that will take care of the y's in your expression

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5.) -14r^2-11rs+15s^2

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6.) x^4-x^2-30

this is a quadratic in $\displaystyle x^2$, factor it as you would a quadratic. but instead of $\displaystyle x$ you are using $\displaystyle x^2$

if it is hard for you to visualize, let $\displaystyle y = x^2$, then you have $\displaystyle y^2 - y - 30$. factorize this, and then replace all the $\displaystyle y$'s with $\displaystyle x^2$'s - Sep 17th 2008, 04:29 PMepetrikSTILL trying to factor!
- Sep 17th 2008, 04:35 PMJhevon
for 2, recall that $\displaystyle a^2 - b^2 = (a + b)(a - b)$

what is your $\displaystyle a$ and $\displaystyle b$ in #2?

as for the others, can you factorize the ones i gave? namely, $\displaystyle 12x^2 - 5x - 3$ (you can use the AC-method, please tell me you know what that is :D) and $\displaystyle y^2 - y - 30$ (remember how to deal with these guys? think of two numbers that you can multiply to get -30 (the constant term) and add to get -1 (the coefficient of y) then put them in the positions of the *'s as in (y + *)(y + *)) - Sep 17th 2008, 05:01 PMepetrik
THANK YOU THANK YOU THANK YOU!!!!!!!

I totally forgot about the AC Method! (Rofl)