I am so confused on these!!
I get stated and then don't know where to start and where to end. Please help.
6a-3a^2-2ab+a^2b
I started with 3(3a-a^2) but not even sure if that is correct.
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
$\displaystyle a(2-a)(3-b) \ \ or \ \ a(a-2)(b-3)$
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.
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
My final answer was 2y(y-2)(y-7)=0
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
Your answers are correct. y = {0, 2, 7}
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.
$\displaystyle 2y[(y^2-7y)-(2y-14)]$
$\displaystyle 2y[y(y-7)-2(y-7)]$
$\displaystyle 2y(y-7)(y-2)=0$
$\displaystyle 2y=0 \ \ or \ \ y-7=0 \ \ or \ \ y-2=0$
$\displaystyle y=0 \ \ or \ \ y=7 \ \ or \ \ y=2$