SOLUTION: Factor the Polynomials: x^3+y^3-x^2y-xy^2 81x^4-16y^4 x^3+2x^2-255x

Question 175687: Factor the Polynomials:
x^3+y^3-x^2y-xy^2
81x^4-16y^4
x^3+2x^2-255x

Answer by srcedwards(3) (Show Source): You can put this solution on YOUR website!
1) Factor by grouping:
x^3 + y^3 - x^2y - xy^2 =
(x^3 + y^3) + -(x^2y + xy^2). The first is a sum of cubes:
[(x + y)(x^2 - xy + y^2)] + [-xy(x + y)]. They both have an x + y in common
(x + y)[(x^2 - xy + y^2) + (-xy)]. Now simplify:
(x + y ( x^2 - 2xy + y^2)
2) This is a difference of squares:
81x^4 - 16y^4 =
(9x^2)^2 - (4y^2)^2 = (9x^2 - 4y^2)(9x^2 + 4y^2). But the first is a difference of squares, so that can be factored again:
(3x - 2y)(3x + 2y)(9x^2 + 4y^2)
3) Factor out a common factor, x:
x^3 + 2x^2 -255x =
x(x^2 + 2x - 255). Can this be factored again? 255 = 15 * 17 which has a difference of two (the middle coefficient), so it can be factored:
x(x - 15)(x - 17)
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