-3x^2+3x+2=0 how do you factor it?..... I know the answer becomes (1-x)(3X+2) but im not sure how to get there

Question

mertsj · Answer

Multiply by -1 to get:
3x^2-3x-2=0
and it is not factorable.

whpalmer4 · Answer

$$(1-x)(3x+2) = 3x+2-3x^2-2x = -3x^2+x+2$$Did you copy part of the problem wrong?

mertsj · Answer

No he didn't copy it wrong.  He forgot to tell you that the right side of the equation is -4x and that the problem has already been solved in another post.

mertsj · Answer

http://openstudy.com/study#/updates/51368d26e4b093a1d94a932b

whpalmer4 · Answer

If you had $-3x^2+x+2=0$ and needed to factor it:

First I would eliminate the annoying - sign on the leading term by multiplying through by -1 as suggested.

$$3x^2-x-2=0$$Your highest power is $x^2$ so you'll have two binomials of the form $(ax+b)$ to multiply together to get your polynomial.
$$3x^2-x-2 = (ax+b)(cx+d)$$
The only factors of 3 are 1 and 3, so we'll take a = 3 and c = 1
$$(3x+b)(x+d) = 3x^2+3dx+bx+bd = 3x^2+(3d+b)x + bd$$
Now we need two numbers b,d, that multiply to -2 and if put in (3d+b) = -1 so that our coefficients match.  Our only two choices for the multiplication are -1,2 and -2,1.  3(-1)+2=-1, 3(-2)+1=-5, so d=-1 and b = 2.  That gives us
$$(3x+2)(x+(-1)) = -1(3x+2)(x-1)$$ as our factoring, after we restore the factor of -1 we dropped.  We could also write it as $$(3x+2)(-x+1) = (3x+2)(1-x)$$if we multiply the (x-1) term by that leading -1.