SOLUTION: Factor completely ...not sure even how to go about doing this problem.... 8x(5th power)y³-6x(4th power)y(5th power)+12x²y³

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Factor completely ...not sure even how to go about doing this problem.... 8x(5th power)y³-6x(4th power)y(5th power)+12x²y³       Log On


   



Question 235500: Factor completely
...not sure even how to go about doing this problem....
8x(5th power)y³-6x(4th power)y(5th power)+12x²y³

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Factor completely
...not sure even how to go about doing this problem....
8x(5th power)y³-6x(4th power)y(5th power)+12x²y³
-----
8x^5y^3 - 6x^4y^5 + 12x^2y^3
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Common Factor: 2x^2y^3
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= 2x^2y^3(4x^3 - 3x^2y^2 + 6)
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Cheers,
Stan H.