# factoring polynomials

• Mar 17th 2011, 06:02 PM
reallylongnickname
factoring polynomials
Im looking to for the steps to solve this: 3(x^2 -2)² (2x-3)^2

apparent answer: 12x^6 - 36x^5 - 21x^4 +144x^3 - 60x^2 - 144x + 108
• Mar 17th 2011, 06:04 PM
pickslides
Is this expression equal to anything?

$\displaystyle 3(x^2-2)^2(2x-3)^2= 0$

Or maybe something like this?

$\displaystyle 3(x^2-2)^2+(2x-3)^2= 0$
• Mar 17th 2011, 06:09 PM
reallylongnickname
No, there is no = 0
• Mar 17th 2011, 06:21 PM
pickslides
So you are looking to expand this? The word solve implies there is an equals sign somewhere in the problem.

$\displaystyle 3(x^2-2)^2(2x-3)^2$

This is the same as

$\displaystyle 3(x^2-2)(x^2-2)(2x-3)(2x-3)$

Now use the distributive law in pairs $\displaystyle (a+b)(c+d) = ac+ad+bc+bd$
• Mar 17th 2011, 06:23 PM
DrSteve
If it's not equal to anything, then there is nothing to solve. Perhaps you are being asked to expand the expression. In this case you just need to do some multiplication using the distributive property several times.

To start you off, $(2x-3)^2=(2x-3)(2x-3)=4x^2-6x-6x+9=4x^2-12x+9$.
• Mar 17th 2011, 07:10 PM
reallylongnickname
for the 3(x^2 - 2)^2 = 3(x^2 - 2) (x^2 - 2) would be 3(x^4 - 4x^2 - 4) then distributed would be 3x^4 -12^2 - 12 but if I collect like terms, I do NOT come up with the correct answer. I must not be applying a correct rule.
• Mar 17th 2011, 07:19 PM
topsquark
Quote:

Originally Posted by reallylongnickname
for the 3(x^2 - 2)^2 = 3(x^2 - 2) (x^2 - 2) would be 3(x^4 - 4x^2 - 4) then distributed would be 3x^4 -12^2 - 12 but if I collect like terms, I do NOT come up with the correct answer. I must not be applying a correct rule.

Leave the 3 out for now. It just makes the coefficients ugly. Okay, so you've got $(x^2 - 2)^2 = x^4 - 4x^2 + 4$ and $(2x - 3)^2 = 4x^2 - 12x + 9$. Now you need to do

$(x^2 - 2)^2(2x - 3)^2 = (x^4 - 4x^2 + 4)(4x^2 - 12x + 9)$.

-Dan
• Mar 17th 2011, 07:54 PM
reallylongnickname
Ahhh I see. The 3 is the last thing to multiply out. Thx topsquark, Dr Steve, and pickslides. I need to stay closer to rules and laws and maybe technique.
• Mar 20th 2011, 12:37 PM
reallylongnickname
New Q. Simplify rational expressions
Ok here is a new Q. (Simplify rational expressions)
(5x^3 -3x +4) + (x^3 -5x^2 -9) I know to first work out each side seperately.
• Mar 20th 2011, 12:42 PM
skeeter
Quote:

Originally Posted by reallylongnickname
Ok here is a new Q. (Simplify rational expressions)
(5x^3 -3x +4) + (x^3 -5x^2 -9) I know to first work out each side seperately.