SOLUTION: Does anyone have a easy way to factor? Her is a problem I just don't know how to do. 9ax^3+15ax^2-14ax

Question 582099: Does anyone have a easy way to factor? Her is a problem I just don't know how to do.
9ax^3+15ax^2-14ax
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Does anyone have a easy way to factor? Her is a problem I just don't know how to do.
9ax^3+15ax^2-14ax
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There is no easy way. You find the combination that works.
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In this one, all 3 terms have an a & and x. Factor those out 1st.
That leaves 9x^2 + 15x - 14
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It'll be
(ax + b)*(cx + d)
a*b = 9, so that's either 1*9 or 3*3
b*d = 14, so that's either 1*14 or 2*7 with different signs (one +, one minus)
Then you find a combination that gives +15x

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
9ax^3+15ax^2-14ax
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= ax(9x^2 + 15x - 14
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= ax(9x^2+21x-6x-14)
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= ax(3x(3x+7)-2(3x+7))
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= ax(3x+7)(3x-2)
===========================================
Cheers,
Stan H.
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