# 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 \$\displaystyle 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 \$\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

Quote:

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

Quote:

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

same advice as the last problem

treat this as \$\displaystyle (x^2)^2 - (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, 04:35 PM
Jhevon
Quote:

Originally Posted by epetrik
I totally appreciate the previous help...but I still don't get number 2, 4,5,6