document.write( "Question 149911: Preform the indicated divison\r
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\n" ); document.write( "\n" ); document.write( "x4+4x3+6x2+4x+1
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\n" ); document.write( " x+1\r
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\n" ); document.write( "\n" ); document.write( "the numbers after the x such as 4x3 is 4x to the third power
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Algebra.Com's Answer #110006 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "Let's simplify this expression using synthetic division\r
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\n" ); document.write( "\n" ); document.write( "Start with the given expression \"%28x%5E4+%2B+4x%5E3+%2B+6x%5E2+%2B+4x+%2B+1%29%2F%28x%2B1%29\"\r
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\n" ); document.write( "\n" ); document.write( "First lets find our test zero:\r
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\n" ); document.write( "\n" ); document.write( "\"x%2B1=0\" Set the denominator \"x%2B1\" equal to zero\r
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\n" ); document.write( "\n" ); document.write( "\"x=-1\" Solve for x.\r
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\n" ); document.write( "\n" ); document.write( "so our test zero is -1\r
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\n" ); document.write( "\n" ); document.write( "Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.\n" ); document.write( "\n" ); document.write( "
-1|14641
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\n" ); document.write( "\n" ); document.write( "Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)\r
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-1|14641
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1
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\n" ); document.write( "\n" ); document.write( " Multiply -1 by 1 and place the product (which is -1) right underneath the second coefficient (which is 4)\r
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-1|14641
|-1
1
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\n" ); document.write( "\n" ); document.write( " Add -1 and 4 to get 3. Place the sum right underneath -1.\r
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-1|14641
|-1
13
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\n" ); document.write( "\n" ); document.write( " Multiply -1 by 3 and place the product (which is -3) right underneath the third coefficient (which is 6)\r
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-1|14641
|-1-3
13
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\n" ); document.write( "\n" ); document.write( " Add -3 and 6 to get 3. Place the sum right underneath -3.\r
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-1|14641
|-1-3
133
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\n" ); document.write( "\n" ); document.write( " Multiply -1 by 3 and place the product (which is -3) right underneath the fourth coefficient (which is 4)\r
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-1|14641
|-1-3-3
133
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\n" ); document.write( "\n" ); document.write( " Add -3 and 4 to get 1. Place the sum right underneath -3.\r
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-1|14641
|-1-3-3
1331
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\n" ); document.write( "\n" ); document.write( " Multiply -1 by 1 and place the product (which is -1) right underneath the fifth coefficient (which is 1)\r
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-1|14641
|-1-3-3-1
1331
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\n" ); document.write( "\n" ); document.write( " Add -1 and 1 to get 0. Place the sum right underneath -1.\r
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-1|14641
|-1-3-3-1
13310
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\n" ); document.write( "\n" ); document.write( "Since the last column adds to zero, we have a remainder of zero. This means \"x%2B1\" is a factor of \"x%5E4+%2B+4x%5E3+%2B+6x%5E2+%2B+4x+%2B+1\"\r
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\n" ); document.write( "\n" ); document.write( "Now lets look at the bottom row of coefficients:\r
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\n" ); document.write( "\n" ); document.write( "The first 4 coefficients (1,3,3,1) form the quotient\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E3+%2B+3x%5E2+%2B+3x+%2B+1\"\r
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\n" ); document.write( "\n" ); document.write( "So \"%28x%5E4+%2B+4x%5E3+%2B+6x%5E2+%2B+4x+%2B+1%29%2F%28x%2B1%29=x%5E3+%2B+3x%5E2+%2B+3x+%2B+1\"\r
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