document.write( "Question 977232: p(x)=x^4-2x^3-8x+16
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Algebra.Com's Answer #598775 by Edwin McCravy(20059) $\"\"$ $\"About$

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p(x)=x^4-2x^3-8x+16
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document.write( "p(x) = x⁴-2x³-8x+16\r\n" );
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document.write( "Out of the first two terms on the right, factor out x³\r\n" );
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document.write( "p(x) = x³(x-2)-8x+16\r\n" );
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document.write( "Out of the last two terms on the right, factor out -8,\r\n" );
document.write( "being careful to remember that when you factor a negative\r\n" );
document.write( "out of a positive, -8 out of +16, you get a negative inside \r\n" );
document.write( "the parentheses:\r\n" );
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document.write( "p(x) = x³(x-2)-8(x-2)\r\n" );
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document.write( "Now there is a common factor of (x-2) in both terms so we\r\n" );
document.write( "factor (x-2) out of both terms:\r\n" );
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document.write( "p(x) = (x-2)(x³-8)\r\n" );
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document.write( "Now since 8 = 2³, we have\r\n" );
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document.write( "p(x) = (x-2)(x³-2³)\r\n" );
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document.write( "And now we use the rule for factoring the difference of two cubes:\r\n" );
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document.write( "p(x) = (x-2)(x-2)(x²+2x+2²)\r\n" );
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document.write( "p(x) = (x-2)²(x²+2x+4)\r\n" );
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document.write( "The only real zero is found by setting factor x-2 equal 0\r\n" );
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document.write( "x-2 = 0\r\n" );
document.write( "  x = 2\r\n" );
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document.write( "It has multiplicity 2 because p(x) has two factors (x-2)(x-2) which \r\n" );
document.write( "when set = 0, provide the zero 2.\r\n" );
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document.write( "There are also two imaginary zeros which we find by setting the other\r\n" );
document.write( "factor x²+2x+4 equal to 0 and using the quadratic formula since it\r\n" );
document.write( "does not factor:\r\n" );
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document.write( "x²+2x+4 = 0\r\n" );
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document.write( "So there are three zeros, 1 real solution of 2 which has multiplicity 2,\r\n" );
document.write( "and two conjugate imaginary or complex zeros  and .\r\n" );
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document.write( "Edwin

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