SOLUTION: Factoring a quadratic with leading coefficient greater than 1 3y^2+7y-6

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Question 295782: Factoring a quadratic with leading coefficient greater than 1
3y^2+7y-6
Found 2 solutions by Fombitz, richwmiller:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
It makes it a little more challenging but the method is the same.
3y%5E2%2B7y-6=%283y%2Ba%29%28y%2Bb%29
If you FOIL the right hand side you get,
%283y%2Ba%29%28y%2Bb%29=3y%5E2%2B3by%2Bay%2Bab
%283y%2Ba%29%28y%2Bb%29=3y%5E2%2B%283b%2Ba%29y%2Bab
So again look for factor of -6:
(1,-6)
(-1,6)
(2,-3)
(-2,3)
and try to solve for the y-coefficient until you hit the right one.
a=1, b=-6
3b%2Ba=7
3%28-6%29%2B1=7
-18%2B1=7
-17=7
No
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.
.
a=-6, b=1
3b%2Ba=7
3-6=7
-3=7
No
.
.
.
a=-1,b=6 and a=6,b=-1 will give similar results.
.
.
.
+a=2, b=-3
3b%2Ba=7
3%28-3%29%2B2=7
-9%2B2=7
-7=7
No, but we only need to change the signs of a and b for this combination to work.
a=-2, b=3.
+3y%5E2%2B7y-6=%283y-2%29%28y%2B3%29+

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
We want factors of -18 which add up to 7
9 and -2
3y^2+9y-2y-6
3y(y+3)-2(y+3)
(3y-2)*(y+3)