document.write( "Question 175687: Factor the Polynomials:\r
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document.write( "x^3+y^3-x^2y-xy^2\r
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document.write( "81x^4-16y^4\r
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document.write( "x^3+2x^2-255x\r
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Algebra.Com's Answer #130936 by srcedwards(3)![]() ![]() ![]() You can put this solution on YOUR website! 1) Factor by grouping:\r \n" ); document.write( "\n" ); document.write( "x^3 + y^3 - x^2y - xy^2 = \n" ); document.write( "(x^3 + y^3) + -(x^2y + xy^2). The first is a sum of cubes: \n" ); document.write( "[(x + y)(x^2 - xy + y^2)] + [-xy(x + y)]. They both have an x + y in common \n" ); document.write( "(x + y)[(x^2 - xy + y^2) + (-xy)]. Now simplify: \n" ); document.write( "(x + y ( x^2 - 2xy + y^2)\r \n" ); document.write( "\n" ); document.write( "2) This is a difference of squares:\r \n" ); document.write( "\n" ); document.write( "81x^4 - 16y^4 = \n" ); document.write( "(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:\r \n" ); document.write( "\n" ); document.write( "(3x - 2y)(3x + 2y)(9x^2 + 4y^2)\r \n" ); document.write( "\n" ); document.write( "3) Factor out a common factor, x:\r \n" ); document.write( "\n" ); document.write( "x^3 + 2x^2 -255x = \n" ); document.write( "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: \n" ); document.write( "x(x - 15)(x - 17) \n" ); document.write( " |