Question 168028: Factor completely by grouping
27x^2 - 117x - 90
Found 2 solutions by gonzo, stanbon:
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! 27x^2 - 117x - 90 factors out to be (3x-15)*(9x+6)
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your c factor possibilities were: +/- [ (90,1), (9,10), (6*15) ]
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your a factor possibilities were: +/- [ (27,1), (9,3) ]
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since your factored equation will be in the form of (a1*x +c1) * (a2*x +c2), you need to find a combination that:
a1*c2 + a2*c1 = -117
c1*c2 = -90
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i tried different combinations until i found the right one.
let a1 = 3
let a2 = 9
let c1 = -15
let c2 = 6
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a1*c2 = 3*6 = 18
a2*c1 = 9*(-15) = -135
a1*c2 + a2*c1 = 18 + (-135) = 18 - 135 = -117
c1*c2 = (-15)*6 = -90
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before i even bothered, i used the part of the quadratic formula sqrt(b^2-4ac) to determine if i had integer roots or not.
if the square root of b^2 - 4ac is an integer, then the quadratic equation has integer roots.
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in the formula above,
a = 27
b = -117
c = -90
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b^2 = 13689
4ac = -9720
sqrt (b^2 - 4ac) = sqrt (13689 - (-9720) = sqrt (13689 + 9720 = sqrt (234098) = 153.
the square root is an integer and so the quadratic equation has integer roots.
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then i went looking for the roots as above.
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Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Factor completely by grouping
27x^2 - 117x - 90
= 9(3x^2 - 13x - 10)
= 9(3x^2 - 15x + 2x - 10)
= 9(3x(x-5)+2(x-5))
= 9(x-5)(3x+2)
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Cheers,
Stan H.
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