# Fatoring polynomials - what am I doing wrong?

• Mar 11th 2011, 06:44 PM
Eloise1995
Fatoring polynomials - what am I doing wrong?
Hi! Here is the problem I was trying to solve, and what I did:

3x^2+7x-6=0
Since 3*(-6)=-18, I need two numbrers, the product of which is -18, and the sum of which is 7. I chose -9 and 2.
3x^2-9x+2x-6=0
3x(x-3)+2(x-3)=0
(3x+2)(x-3)=0
3x+2=0 or x-3=0
x=3 x=-2/3

However, the back of the textbook gives the answer x=-3 x=2/3
What is wrong? Thanks if you could help, big time! I didn't do grade 10 math, and in grade 11 they only review factoring for a couple of lessons.

P.S. Sorry for doubleposting!!
• Mar 11th 2011, 06:52 PM
pickslides
I see it like this

\$\displaystyle \displaystyle 3x^2+7x-6=0\$

\$\displaystyle \displaystyle 3x^2+9x-2x-6=0\$

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

\$\displaystyle \displaystyle (x+3)(3x-2)=0\$
• Mar 11th 2011, 06:56 PM
Prove It
\$\displaystyle \displaystyle -9 + 2 \neq 7\$. But \$\displaystyle \displaystyle 9 + (-2)\$ does...