# I'm totally confused!

• Dec 15th 2008, 09:21 AM
tpoma
I'm totally confused!
I am so confused on these!!

6a-3a^2-2ab+a^2b

I started with 3(3a-a^2) but not even sure if that is correct. (Headbang)

Thank you
• Dec 15th 2008, 09:34 AM
running-gag
Hi

\$\displaystyle 6a-3a^2-2ab+a^2b = 2a(3-b)+a^2(b-3) = (a^2-2a)(b-3) = a(a-2)(b-3)\$
• Dec 15th 2008, 09:35 AM
masters
Quote:

Originally Posted by tpoma
I am so confused on these!!

6a-3a^2-2ab+a^2b

I started with 3(3a-a^2) but not even sure if that is correct. (Headbang)

Thank you

First, factor out an a, then try gouping the terms when you have a 4-term polynomial.

Group the first 2, then the second 2 and pull out common factors.

\$\displaystyle 6a-3a^2-2ab+a^2b\$

\$\displaystyle a(6-3a-2b+ab)\$

\$\displaystyle a [ (6-3a)-(2b-ab) ] \$

\$\displaystyle a [ 3(2-a)-b(2-a) ] \$

How's this working out for you? Can you finish up?

I'll finish up since running-gag had a slightly different approach

\$\displaystyle a(2-a)(3-b) \ \ or \ \ a(a-2)(b-3)\$
• Dec 15th 2008, 10:42 AM
tpoma
Ok, I am trying this again, can you please tell me if this is correct or anywhere close to what I am trying to accomplish? Thanks

4k-8-k^3+2k^2

I have come up with 2k(2+k^2+2k) I don't feel confident that it is anywhere close to what I am suppose to have.
• Dec 15th 2008, 11:39 AM
running-gag
No (Worried)

\$\displaystyle 4k-8-k^3+2k^2 = 4(k-2) - k^2(k-2) = (4 - k^2)(k-2)\$

\$\displaystyle 4k-8-k^3+2k^2 = (2+k)(2-k)(k-2) = -(k+2)(k-2)^2\$
• Dec 15th 2008, 01:12 PM
tpoma
Quick question,

On this problem why is it that the 9 seems to disapear? It seems to just drop off????

2y^3-18y^2= -28y

2y=0 y-2=0
y=0 y=2

y-7=0 y=7
Where did the 9 go to?? I don't understand it but I think I got it right.

Thanks
• Dec 15th 2008, 01:49 PM
masters
Quote:

Originally Posted by tpoma
Quick question,

On this problem why is it that the 9 seems to disapear? It seems to just drop off????

2y^3-18y^2= -28y

2y=0 y-2=0
y=0 y=2

y-7=0 y=7
Where did the 9 go to?? I don't understand it but I think I got it right.

Thanks

Here's how it goes:

\$\displaystyle 2y^3-18y^2=-28y\$

\$\displaystyle 2y^3-18y^2+28y=0\$

\$\displaystyle 2y(y^2-9y+14)=0\$

When you factor the trinomial, the sum of the two factors that make 14 must add to make -9.

\$\displaystyle 2y(y-7)(y-2)=0\$

You see. The -9y is made up of -7y + -2y. You may see it better this way.

\$\displaystyle 2y(y^2-7y-2y+14)=0\$

Here, I just replaced the -9y in the middle with the sum of the two factors that made +14. Now, if I group the 4 term polynomial, I arrive at the desired factored result.

\$\displaystyle 2y[(y^2-7y)-(2y-14)]\$

\$\displaystyle 2y[y(y-7)-2(y-7)]\$

\$\displaystyle 2y(y-7)(y-2)=0\$

\$\displaystyle 2y=0 \ \ or \ \ y-7=0 \ \ or \ \ y-2=0\$

\$\displaystyle y=0 \ \ or \ \ y=7 \ \ or \ \ y=2\$
• Dec 15th 2008, 01:58 PM
tpoma
ahhhh got ya, so I could have used FOIL on this problem once I got the equation set up....didn't even see it in a FOIL format! Thank you for your explination! They always seem to help when I am having problems!! (Clapping)