Asked 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 second-13x=6
-5x to the second-6x=1
-and xto the second + 6x=10
I can't use the quadratic equation. I have to use the a times c method.
-3x to the second+18x+12
-5x to the second-13x=6
-5x to the second-6x=1
-and xto the second + 6x=10
I can't use the quadratic equation. I have to use the a times c method.
Answers
Answered by
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(x-3) + 2(x-3) = 0
now x-3 is a common factor, so ...
(x-3)(5x+2) = 0
x-3 = 0 or 5x+2 = 0
x = 3 or x = -2/5
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(x-3) + 2(x-3) = 0
now x-3 is a common factor, so ...
(x-3)(5x+2) = 0
x-3 = 0 or 5x+2 = 0
x = 3 or x = -2/5
Answered by
Reiny
Do the others the same way
Answered by
John
So You using the quadratic or a time c method will give me the same answer?
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