document.write( "Question 119164: I'm learning to divide polynomials by polynomials. I have been able to work several problems, but then get stuck as they progress. For example:\r
\n" ); document.write( "\n" ); document.write( "25x to the 5th power take away x to the third power take away 8x take away 2x square divided by 5xsquare take away 4x. I have placed the dividend and divisor in descending order and even tried adding 0x to the fourth power in the dividend. I seem to get the first half correct, but can't get the latter half. What am I doing wrong? \r
\n" ); document.write( "\n" ); document.write( "Jennifer
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Algebra.Com's Answer #87273 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
You have done the first steps correctly, that is you put the terms in decending order of degree and put a zero coefficient placeholder in for the missing 4th degree term. I presume you ended up with:\r
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\n" ); document.write( "\n" ); document.write( "\"%2825x%5E5%2B0x%5E4-x%5E3-2x%5E2-8x%29%2F%285x%5E2-4x%29\"\r
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\n" ); document.write( "\n" ); document.write( "The next step is to factor the denominator. \"5x%5E2-4x=x%285x-4%29\". Now your problem looks like this:\r
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\n" ); document.write( "\n" ); document.write( "\"%2825x%5E5%2B0x%5E4-x%5E3-2x%5E2-8x%29%2Fx%285x-4%29\"\r
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\n" ); document.write( "\n" ); document.write( "But notice that there is an \"x\" in every term of the numerator which means you can divide through by the \"x\" in the denominator to get:\r
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\n" ); document.write( "\n" ); document.write( "\"%2825x%5E4%2B0x%5E3-x%5E2-2x-8%29%2F%285x-4%29\"\r
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\n" ); document.write( "\n" ); document.write( "Now your polynomial long division shouldn't be so ugly. I don't know how to render the process on this site, so I'll just talk you through it. Write back if you have trouble understanding.\r
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\n" ); document.write( "\n" ); document.write( "Step 1: \"5x\" goes into \"25x%5E4\" \"5x%5E3\" times, so \"5x%5E3\" is the first term of your quotient polynomial.\r
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\n" ); document.write( "\n" ); document.write( "Step 2: \"5x%5E3\" times \"-4\" is \"-20x%5E3\" and \"5x%5E3\" times \"5x\" is \"25x%5E4\", so \"25x%5E4-20x%5E3\" is your first partial product.\r
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\n" ); document.write( "\n" ); document.write( "Step 3: Subtract \"25x%5E4-20x%5E3\" from the first two terms of the dividend polynomial giving you \"0x%5E4%2B20x%5E3\".\r
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\n" ); document.write( "\n" ); document.write( "Step 4: Bring down the next term, \"-x%5E2\" to form your next partial dividend \"20x%5E3-x%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "Step 6: \"5x\" goes into \"20x%5E3\" \"4x%5E2\" times, so this is the next term of your quotient polynomial. So far, your quotient should look like \"5x%5E3%2B4x%5E2\".\r
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\n" ); document.write( "\n" ); document.write( "Step 7: \"4x%5E2\" times \"-4\" is \"-16x%5E2\" and \"4x%5E2\" times \"5x\" is \"20x%5E3\", so \"20x%5E3-16x%5E2\" is your next partial product.\r
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\n" ); document.write( "\n" ); document.write( "Step 8: Subtract this from your previous partial dividend, \"20x%5E3-x%5E2\" (remembering to change the sign and add when you subtract), to get \"0x%5E3%2B15x%5E2\".\r
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\n" ); document.write( "\n" ); document.write( "Step 9: Bring down the next term of the dividend, \"-2x\", to form your next partial dividend, \"15x%5E2-2x\".\r
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\n" ); document.write( "\n" ); document.write( "Step 10: \"5x\" goes into \"15x%5E2\" \"3x\" times giving you the next term of your quotient. Your quotient should now look like \"5x%5E3%2B4x%5E2%2B3x\".\r
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\n" ); document.write( "\n" ); document.write( "Step 11: \"3x\" times \"-4\" is \"-12x\" and \"3x\" times \"5x\" is \"15x%5E2\", so \"15x%5E2-12x\" is your next partial product.\r
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\n" ); document.write( "\n" ); document.write( "Step 12: Subtract this from your previous partial dividend, \"15x%5E2-2x\", resulting in \"0x%5E2%2B10x\"\r
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\n" ); document.write( "\n" ); document.write( "Step 13: Bring down the last term, \"-8\" to form your next partial dividend \"10x-8\"\r
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\n" ); document.write( "\n" ); document.write( "Step 14: \"5x\" goes into \"10x\" \"2\" times, so \"2\" is the last term of your quotient which should look like \"5x%5E3%2B4x%5E2%2B3x%2B2\".\r
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\n" ); document.write( "\n" ); document.write( "Step 15: \"2\" times \"-4\" is \"-8\" and \"2\" times \"5x\" is \"10x\" so your next partial product is \"10x-8\".\r
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\n" ); document.write( "\n" ); document.write( "Step 16: Subtract. \"%2810x-8%29-%2810x-8%29=0\". There is no remainder, so you are done.\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps.
\n" ); document.write( "John
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