document.write( "Question 134593: Factor each polynomial completely, given that the binomial following it is a factor of the polynomial.
\n" ); document.write( "x^3+2x^2-5x-6, x+3\r
\n" ); document.write( "\n" ); document.write( "Elementary and Intermediate Algebra
\n" ); document.write( "Chapter 5 - Factoring
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Algebra.Com's Answer #98460 by solver91311(24713)\"\" \"About 
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Use polynomial long division to divide \"x%2B3\" into \"x%5E3%2B2x%5E2-5x-6\"\r
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\n" ); document.write( "\n" ); document.write( "\"x\" goes into \"x%5E3\" \"x%5E2\" times, so \"x%5E2\" is the first term of the quotient.\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2\" times \"x%2B3\" is \"x%5E3%2B3x%5E2\".\r
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\n" ); document.write( "\n" ); document.write( "Subtract \"x%5E3%2B3x%5E2\" from \"x%5E3%2B2x%5E2\" and get \"-x%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "Bring down the \"-5x\" to form \"-x%5E2-5x\"\r
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\n" ); document.write( "\n" ); document.write( "\"x\" goes into \"-x%5E2\" \"-x\" times, so \"-x\" becomes the second term of the quotient.\r
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\n" ); document.write( "\n" ); document.write( "\"-x\" times \"x%2B3\" is \"-x%5E2-3x\"\r
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\n" ); document.write( "\n" ); document.write( "Subtract \"-x%5E2-3x\" from \"-x%5E2-5x\" and get \"-2x\"\r
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\n" ); document.write( "\n" ); document.write( "Bring down the \"-6\" to form \"-2x-6\"\r
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\n" ); document.write( "\n" ); document.write( "\"x\" goes into \"-2x\" \"-2\" times, so \"-2\" becomes the third term of the quotient.\r
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\n" ); document.write( "\n" ); document.write( "\"-2\" times \"x%2B3\" is \"-2x-6\"\r
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\n" ); document.write( "\n" ); document.write( "Subtract \"-2x-6\" from \"-2x-6\" and get 0. So you have verified that \"x%2B3\" is a factor. Now put the quotient together:\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2-x-2\"\r
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\n" ); document.write( "\n" ); document.write( "\"-2%2A1=-2\" and \"-2%2B1=-1\", so the trinomial quotient factors to:\r
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\n" ); document.write( "\n" ); document.write( "\"%28x-2%29%28x-1%29\"\r
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\n" ); document.write( "\n" ); document.write( "Therefore: \"x%5E3%2B2x%5E2-5x-6=%28x%2B3%29%28x-2%29%28x-1%29\"
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