document.write( "Question 129815: I need help with this question\r
\n" ); document.write( "\n" ); document.write( "Use synthetic division to find the remainder
\n" ); document.write( "(x3-3x+10 divide (x-2)
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Algebra.Com's Answer #94845 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%5E3+-+3x+%2B+10%29%2F%28x-2%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-2=0\" Set the denominator \"x-2\" equal to zero\r
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\n" ); document.write( "\n" ); document.write( "\"x=2\" Solve for x.\r
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\n" ); document.write( "\n" ); document.write( "so our test zero is 2\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.(note: remember if a polynomial goes from \"1x%5E3\" to \"-3x%5E1\" there is a zero coefficient for \"x%5E2\". This is simply because \"x%5E3+-+3x+%2B+10\" really looks like \"1x%5E3%2B0x%5E2%2B-3x%5E1%2B10x%5E0\"\n" ); document.write( "\n" ); document.write( "
2|10-310
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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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2|10-310
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1
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\n" ); document.write( "\n" ); document.write( " Multiply 2 by 1 and place the product (which is 2) right underneath the second coefficient (which is 0)\r
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2|10-310
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1
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\n" ); document.write( "\n" ); document.write( " Add 2 and 0 to get 2. Place the sum right underneath 2.\r
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2|10-310
|2
12
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\n" ); document.write( "\n" ); document.write( " Multiply 2 by 2 and place the product (which is 4) right underneath the third coefficient (which is -3)\r
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2|10-310
|24
12
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\n" ); document.write( "\n" ); document.write( " Add 4 and -3 to get 1. Place the sum right underneath 4.\r
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2|10-310
|24
121
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\n" ); document.write( "\n" ); document.write( " Multiply 2 by 1 and place the product (which is 2) right underneath the fourth coefficient (which is 10)\r
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2|10-310
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121
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\n" ); document.write( "\n" ); document.write( " Add 2 and 10 to get 12. Place the sum right underneath 2.\r
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2|10-310
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12112
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\n" ); document.write( "\n" ); document.write( "Since the last column adds to 12, we have a remainder of 12. This means \"x-2\" is not a factor of \"x%5E3+-+3x+%2B+10\"\r
\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 3 coefficients (1,2,1) form the quotient\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2+%2B+2x+%2B+1\"\r
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\n" ); document.write( "\n" ); document.write( "and the last coefficient 12, is the remainder, which is placed over \"x-2\" like this\r
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\n" ); document.write( "\n" ); document.write( "\"12%2F%28x-2%29\"\r
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\n" ); document.write( "\n" ); document.write( "Putting this altogether, we get:\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2+%2B+2x+%2B+1%2B12%2F%28x-2%29\"\r
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\n" ); document.write( "\n" ); document.write( "So \"%28x%5E3+-+3x+%2B+10%29%2F%28x-2%29=x%5E2+%2B+2x+%2B+1%2B12%2F%28x-2%29\"\r
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\n" ); document.write( "\n" ); document.write( "which looks like this in remainder form:\r
\n" ); document.write( "\n" ); document.write( "\"%28x%5E3+-+3x+%2B+10%29%2F%28x-2%29=x%5E2+%2B+2x+%2B+1\" remainder 12\r
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\n" ); document.write( "\n" ); document.write( "You can use this online polynomial division calculator to check your work\r
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