document.write( "Question 891884: A polynomial P is given. \r
\n" ); document.write( "\n" ); document.write( "P(x) = x^6-7x^3-8
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\n" ); document.write( "(a) Find all zeros of P, real and complex. (Enter your answers as a comma-separated list. If a root has multiplicity greater than one, only enter the root once.)
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\n" ); document.write( "b) Factor P Completely
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Algebra.Com's Answer #540126 by josgarithmetic(39618)\"\" \"About 
You can put this solution on YOUR website!
Possible maximum of six roots. Try synthetic division on \"x%5E6%2B0%2Ax%5E5%2B0%2Ax%5E4-7x%5E3%2B0%2Ax%5E2%2B0%2Ax-8\" and check the possible roots plus-and-minus 1,2,4,8.\r
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\n" ); document.write( "\n" ); document.write( "According to that, roots found: -1, 2
\n" ); document.write( "Those seem to be the only rational roots, leaving as quotient coefficients 1,1,3,-2,4 for the polynomial factor \"x%5E4%2Bx%5E3%2B3x%5E2-2x%2B4\". There are no other real zeros or roots (graphing feature of google was used to see that).\r
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\n" ); document.write( "\n" ); document.write( "Should be some other complex roots. Maybe a substitution for x^3 as another temporary variable may help. If you will do this, then you can factorize this way:\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E6-7x%5E3-8=%28x%5E3%2B1%29%28x%5E3-8%29\".
\n" ); document.write( "You can derive for yourself OR look for sum and difference of cubes in a book to obtain;
\n" ); document.write( "\"highlight_green%28%28x%2B1%29%28x%5E2-x%2B1%29%28x-2%29%28x%5E2%2B2x%2B4%29%29\", and from these, you can both see the roots already found, and can find the other four complex roots using the general solution for a quadratic formula.
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