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January 31, 2015

January 31, 2015

Posted by **wendy** on Thursday, January 11, 2007 at 11:22pm.

6y(to the sixth power) - 5y(to the third power) - 4

6y^6 - 5y^3 -4

= (2y^3 + 1)(3y^3- 4)

I do not believe the cubic terms are separable into quadratic factors.

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