RE: Math
posted by John .
Need help on factoring. HOw exactly would I solve these(big test coming up):
3x to the second+18x+12
5x to the second13x=6
5x to the second6x=1
and xto the second + 6x=10
I can't use the quadratic equation. I have to use the a times c method.

RE: Math 
Reiny
Two things:
1. I think your first equation has a typo, since it is not an equation
2. I think you are using the  in front of each equation not as a negative sign, but rather as a dash. Why????
So I will assume that your second equation is
5x^2  13x = 6 (the ^ means 'to the exponent')
5x^2  13x  6 = 0
(a)(c) = 5(6) = 30
now look for factors of 30 which have a sum of 13
that would be 15 and 2
check :(15)(2) = 30, 15 + 2 = 13
now replace the middle term of 13x with 15x+2x
5x^2  15x + 2x  6 = 0
take out a common factor from the first two terms, and from the last two terms
5x(x3) + 2(x3) = 0
now x3 is a common factor, so ...
(x3)(5x+2) = 0
x3 = 0 or 5x+2 = 0
x = 3 or x = 2/5 
RE: Math 
Reiny
Do the others the same way

RE: Math 
John
So You using the quadratic or a time c method will give me the same answer?
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