document.write( "Question 83397: Q1 (i) Factor the expression
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\n" ); document.write( " x^5 – 2x^4 + x^3 – 2x^2 – 2x + 4 = 0 \n" ); document.write( "
Algebra.Com's Answer #59918 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Here are the 1st and 3rd.\r\n" );
document.write( "I haven't figured out the 2nd one yet.  \r\n" );
document.write( "Is that the way it was given?  No errors?\r\n" );
document.write( "Is it supposed to be a division?\r\n" );
document.write( "You were asked to \"expand a quotient\"?  \r\n" );
document.write( "That doesn't seem right.\r\n" );
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document.write( "Anyway here are the first and third ones:\r\n" );
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document.write( "Q1 (i) Factor the expression                          \r\n" );
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document.write( "\"4y%5E2-12xy-8yz%2B9x%5E2%2B12xz%2B4z%5E2\"                              \r\n" );
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document.write( "Rearrange the terms as\r\n" );
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document.write( "\"9x%5E2-12xy%2B4y%5E2\" + \"12xz+-+8yz\" + \"4z%5E2\"\r\n" );
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document.write( "The first three terms factor as \"%283x-2y%29%283x-2y%29\" or \"%283-2x%29%5E2\"\r\n" );
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document.write( "We can take \"4z\" out of the \r\n" );
document.write( "3rd and 4th terms and we now have\r\n" );
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document.write( "\"%283x-2y%29%5E2+%2B+4z%283x-2y%29+%2B+4z%B2\"\r\n" );
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document.write( "Now let \"w+=+%283x-2y%29\"\r\n" );
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document.write( "Then the above becomes\r\n" );
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document.write( "\"w%5E2+%2B4zw+%2B+4z%5E2\"\r\n" );
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document.write( "which factors as\r\n" );
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document.write( "\"%28w+%2B+2z%29%28w%2B2z%29\" or\r\n" );
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document.write( "\"%28w+%2B+2z%29%5E2\"\r\n" );
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document.write( "Now replace the w by (3x-2y)\r\n" );
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document.write( "\"%283x-2y%2B2z%29%5E2\"\r\n" );
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document.write( "(iii) Solve the equation:\r\n" );
document.write( "      \r\n" );
document.write( "\"x%5E5-2x%5E4%2Bx%5E3-+2x%5E2+-+2x+%2B+4+=+0\" \r\n" );
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document.write( "We notice that the coefficients of the\r\n" );
document.write( "1st, 3rd, and 5th terms are 1,1,-2, and that\r\n" );
document.write( "the coefficients of the 2nd, 4th, and 6th\r\n" );
document.write( "terms are -2,-2,4 and that these are\r\n" );
document.write( "proportional, so we rearrange the terms:\r\n" );
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document.write( "\"x%5E5%2Bx%5E3-2x-2x%5E2-2x%2B4=0\"\r\n" );
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document.write( "Factor x out of the first 3 terms and -2\r\n" );
document.write( "out of the last 3 terms:\r\n" );
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document.write( "\"x%28x%5E4%2Bx%5E2-2%29-2%28x%5E4%2Bx%5E2-2%29=0\" \r\n" );
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document.write( "Now we can take out the common factor \r\n" );
document.write( "\"%28x%5E4%2Bx%5E2-2%29\" and we have:\r\n" );
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document.write( "\"%28x%5E4%2Bx%5E2-2%29%28x-2%29=0\"\r\n" );
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document.write( "Now we can factor the first parenthetical \r\n" );
document.write( "expression:\r\n" );
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document.write( "\"%28x%5E2%2B2%29%28x%5E2-1%29%28x-2%29=0\"\r\n" );
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document.write( "and finally factor the expression in \r\n" );
document.write( "the middle parentheses as the difference \r\n" );
document.write( "of two squares:\r\n" );
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document.write( "\"%28x%5E2%2B2%29%28x-1%29%28x%2B1%29%28x-2%29=0\"\r\n" );
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document.write( "Now we set each factor = 0:\r\n" );
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document.write( "Setting first factor = 0\r\n" );
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document.write( "\"x%5E2%2B2=0\"\r\n" );
document.write( "\"x%5E2=-2\"\r\n" );
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document.write( "Use the principle of square roots  \r\n" );
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document.write( "    x = ±\"sqrt%28-2%29\"\r\n" );
document.write( "    x = ±\"i%2Asqrt%282%29\"\r\n" );
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document.write( "Setting second factor = 0\r\n" );
document.write( "    x - 1 = 0 gives x = 1\r\n" );
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document.write( "Setting third factor = 0\r\n" );
document.write( "    x + 1 = 0 gives x = -1\r\n" );
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document.write( "Setting fourth factor = 0\r\n" );
document.write( "    x - 2 = 0 gives x = 2\r\n" );
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document.write( "So the 5 solutions are ±i\"sqrt%282%29\",1,-1, and 2  \r\n" );
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document.write( "Edwin
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