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.Com
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) (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
--------------------------------
Common Factor: 2x^2y^3
--------------------------------
= 2x^2y^3(4x^3 - 3x^2y^2 + 6)
===================================
Cheers,
Stan H.
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