# Factoring! YUCK!

• Sep 17th 2008, 03:20 PM
epetrik
Factoring! 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 PM
Jhevon
Quote:

Originally Posted by epetrik
1.) 36x+18xy-9y-18

factor by grouping. there is a common 18x in the first two terms, and a common -9 in the last two

Quote:

2.) 4-36x^2
write as $2^2 - (6x)^2$

now apply the formula for the difference of two squares factorization

Quote:

3.) 42x^2-24
begin by factoring out the common 6

now, what do you aee?

Quote:

4.) 12x^2-5xy-3y^2
these are one of those two-way quadratic thingys.

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

Quote:

5.) -14r^2-11rs+15s^2
same advice as the last problem

Quote:

6.) x^4-x^2-30
treat this as $(x^2)^2 - (x^2) - 30$

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

if it is hard for you to visualize, let $y = x^2$, then you have $y^2 - y - 30$. factorize this, and then replace all the $y$'s with $x^2$'s
• Sep 17th 2008, 04:29 PM
epetrik
STILL trying to factor!
I totally appreciate the previous help...but I still don't get number 2, 4,5,6

Quote:

Originally Posted by Jhevon
factor by grouping. there is a common 18x in the first two terms, and a common -9 in the last two

write as $2^2 - (6x)^2$

now apply the formula for the difference of two squares factorization

begin by factoring out the common 6

now, what do you aee?

these are one of those two-way quadratic thingys.

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

same advice as the last problem

treat this as $(x^2)^2 - (x^2) - 30$

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

if it is hard for you to visualize, let $y = x^2$, then you have $y^2 - y - 30$. factorize this, and then replace all the $y$'s with $x^2$'s

• Sep 17th 2008, 04:35 PM
Jhevon
Quote:

Originally Posted by epetrik
I totally appreciate the previous help...but I still don't get number 2, 4,5,6
for 2, recall that $a^2 - b^2 = (a + b)(a - b)$
what is your $a$ and $b$ in #2?
as for the others, can you factorize the ones i gave? namely, $12x^2 - 5x - 3$ (you can use the AC-method, please tell me you know what that is :D) and $y^2 - y - 30$ (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 + *))