document.write( "Question 825058: This question has two parts, and I'm kind of confused on what is the difference in the two equations.\r
\n" ); document.write( "\n" ); document.write( "a). Use synthetic division to find the quotient and the remainder when x^4-4x^2+2x+5 is divided by x-2\r
\n" ); document.write( "\n" ); document.write( "b) Use long division to find the quotient and remainder when 2x^5+4x^4-x^3-x^2+7 is divided by 2x^2-1.\r
\n" ); document.write( "\n" ); document.write( "Please help this is a struggle for me. \n" ); document.write( "
Algebra.Com's Answer #497002 by Edwin McCravy(20054)\"\" \"About 
You can put this solution on YOUR website!
This question has two parts, and I'm kind of confused on what is the difference in the two equations.\r
\n" ); document.write( "\n" ); document.write( "a). Use synthetic division to find the quotient and the remainder when x^4-4x^2+2x+5 is divided by x-2
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document.write( "Both problems CAN be done by long division.  I'll do the first\r\n" );
document.write( "one by long division first just to show you that it can be done\r\n" );
document.write( "that way.  Then I'll show you the shortcut, known as \"synthetic\"\r\n" );
document.write( "division.  Here's the long division.  There is no x³ term so we\r\n" );
document.write( "must put in a placeholder +0x³: \r\n" );
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document.write( "            x³ + 2x² +  6x + 14\r\n" );
document.write( "x - 2)x4 + 0x³ + 2x² +  2x +  5\r\n" );
document.write( "      x4 - 2x³\r\n" );
document.write( "           2x³ + 2x²\r\n" );
document.write( "           2x³ - 4x²\r\n" );
document.write( "                 6x² +  2x\r\n" );
document.write( "                 6x² - 12x\r\n" );
document.write( "                       14x +  5\r\n" );
document.write( "                       14x - 28\r\n" );
document.write( "                             33\r\n" );
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document.write( "The quotient is x³+2x²+6x+14 and the remainder is 33\r\n" );
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document.write( "However your teacher doesn't want you to do it that way. He/she wants\r\n" );
document.write( "you to learn the shortcut, synthetic division, where you don't\r\n" );
document.write( "have to write the variables down.  But synthetic division only works \r\n" );
document.write( "when you're dividing by either \"x+↏\" or \"x-↏\", so you can't do the \r\n" );
document.write( "second problem by synthetic division, but you can do the first one, \r\n" );
document.write( "that I just did above by long division.\r\n" );
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document.write( "In long division you have to subtract by thinking of the sign\r\n" );
document.write( "changed and then add.  But in synthetic division, we change the\r\n" );
document.write( "sign of the number after the x, so we can add instead of \r\n" );
document.write( "subtract.  Here we are dividing by x-2, so we change the sign \r\n" );
document.write( "of -2 to +2 and write 2 for the divisor.  Then instead of\r\n" );
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document.write( "1x4+0x³+2x²+2x+5 we just write 1  0  2  2  5 and draw a vertical\r\n" );
document.write( "line between the 2 and the 1, and a horizontal line below like\r\n" );
document.write( "this: \r\n" );
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document.write( "2| 1  0  2  2  5\r\n" );
document.write( " |______________     \r\n" );
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document.write( "Begin by bringing down the 1\r\n" );
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document.write( "2| 1  0  2  2  5\r\n" );
document.write( " |______________\r\n" );
document.write( "   1\r\n" );
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document.write( "Multiply the 1 by the 2 in the upper left, getting 2 and put it \r\n" );
document.write( "above and to the right, above the line underneath the 0:\r\n" );
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document.write( "2| 1  0  2  2  5\r\n" );
document.write( " |    2           \r\n" );
document.write( "   1  \r\n" );
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document.write( "Now we add the 0 and the 2, getting 2, and write it below the\r\n" );
document.write( "line under the 2:\r\n" );
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document.write( "2| 1  0  2  2  5\r\n" );
document.write( " |    2           \r\n" );
document.write( "   1  2  \r\n" );
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document.write( "Multiply the 2 on the bottom line by the 2 in the upper left, \r\n" );
document.write( "getting 4 and put the 4 above and to the right, above the line\r\n" );
document.write( "underneath the 2 in the middle on the top.\r\n" );
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document.write( "2| 1  0  2  2  5\r\n" );
document.write( " |    2  4      \r\n" );
document.write( "   1  2 \r\n" );
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document.write( "Now we add the 2 and the 4 getting 6 and we write the 6 underneath\r\n" );
document.write( "the 4 below the line:\r\n" );
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document.write( "2| 1  0  2  2  5\r\n" );
document.write( " |    2  4      \r\n" );
document.write( "   1  2  6\r\n" );
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document.write( "Multiply the 6 on the bottom line by the 2 in the upper left, \r\n" );
document.write( "getting 12 and put the 12 above and to the right, above the line\r\n" );
document.write( "underneath the next 2 on the top.\r\n" );
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document.write( "2| 1  0  2  2  5\r\n" );
document.write( " |    2  4 12   \r\n" );
document.write( "   1  2  6\r\n" );
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document.write( "Now we add the 2 and the 12 getting 14 and we write the 14 underneath\r\n" );
document.write( "the 12 below the line:\r\n" );
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document.write( "2| 1  0  2  2  5\r\n" );
document.write( " |    2  4 12   \r\n" );
document.write( "   1  2  6 14\r\n" );
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document.write( "Multiply the 14 on the bottom line by the 2 in the upper left, \r\n" );
document.write( "getting 28 and put the 28 above and to the right, above the line\r\n" );
document.write( "underneath the 5 on the top.\r\n" );
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document.write( "2| 1  0  2  2  5\r\n" );
document.write( " |    2  4 12 28\r\n" );
document.write( "   1  2  6 14\r\n" );
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document.write( "Now we add the 5 and the 28 getting 33 and we write the 33 underneath\r\n" );
document.write( "the 28 below the line:\r\n" );
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document.write( "2| 1  0  2  2  5\r\n" );
document.write( " |    2  4 12 28\r\n" );
document.write( "   1  2  6 14 33\r\n" );
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document.write( "Finally we interpret the row of numbers  1  2  6 14 33 on \r\n" );
document.write( "the bottom.  Since the largest power of x in x^4-4x^2+2x+5 \r\n" );
document.write( "is 4, the largest power of x in the quotient will be 1 \r\n" );
document.write( "less than 4, or 3. So all the numbers but the last one are\r\n" );
document.write( "the coefficients of the quotient, so the quotient is\r\n" );
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document.write( "1x³+2x²+6x+14 and the last number 33 is the remainder.\r\n" );
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document.write( "Notice that that is the same answer as when we used long division \r\n" );
document.write( "above.\r\n" );
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\n" ); document.write( "b) Use long division to find the quotient and remainder when 2x5+4x4-x³-x²+7 is divided by 2x²-1.
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document.write( "We must put in a zero-placeholder in both the divisor and the dividend \r\n" );
document.write( "and consider this as dividing 2x5+4x4-x³-x²+0x+7 by 2x²+0x-1.   \r\n" );
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document.write( "This problem involves some zeros as well as some fractions at the end:\r\n" );
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document.write( "                          x³ + 2x² + 0x + 1/2 = quotient\r\n" );
document.write( "2x² + 0x - 1)2x5 + 4x4 -  x³ -  x² + 0x +   7\r\n" );
document.write( "             2x5 + 0x4 -  x³\r\n" );
document.write( "                   4x4 + 0x³ -  x²\r\n" );
document.write( "                   4x4 + 0x³ - 2x²\r\n" );
document.write( "                         0x³ +  x² + 0x\r\n" );
document.write( "                         0x³ + 0x² + 0x\r\n" );
document.write( "                                x² + 0x +    7\r\n" );
document.write( "                                x² + 0x -  1/2\r\n" );
document.write( "                                     0x + 15/2 = remainder   \r\n" );
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document.write( "Notice I got the 15/2 remainder by 7-\"%28-1%2F2%29\" = 7+\"1%2F2\" = \"14%2F2%2B1%2F2\" = \"15%2F2\"\r\n" );
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document.write( "Edwin
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