document.write( "Question 1133349: IN THE PARTIAL FRACTION DECOMPOSITION of x^2/[(x-1)^2 (x+1)^2] the format
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Algebra.Com's Answer #750553 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
IN THE PARTIAL FRACTION DECOMPOSITION of x^2/(x-1)^2 (x+1)^2 the format is as follows:
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document.write( "Multiply through by the LCD:\r\n" );
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\n" ); document.write( "I let x = 1
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document.write( "\"+1+=+B%2A%282%29%5E2\"\r\n" );
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document.write( "\"+1+=+B%2A4\"\r\n" );
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document.write( "\"1%2F4=B\"\r\n" );
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\n" ); document.write( "and got B= 1/4 (which is correct) THEN let x=-1 to get D= 1/4
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document.write( "\"+1+=+D%2A%28-2%29%5E2\"\r\n" );
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document.write( "\"+1+=+D%2A4\"\r\n" );
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document.write( "\"1%2F4=D\"\r\n" );
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\n" ); document.write( "Am stuck at this point Can not get A or C
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document.write( "Go back to:\r\n" );
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document.write( "Substitute the values you got for B and D\r\n" );
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document.write( "Clear of fractions:\r\n" );
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document.write( "Substitute some number you haven't substituted before. \r\n" );
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document.write( "(Note: You don't have to just substitute numbers that make terms equal to zero.\r\n" );
document.write( "You can substitute ANY number for x and it will always give a true equation,\r\n" );
document.write( "even if none of the terms become 0. So the easiest number you haven't\r\n" );
document.write( "substituted is x=0.\r\n" );
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document.write( "\"+0+=+4A%2A%28-1%29%281%29%5E2+%2B+%281%29%5E2+%2B4C%2A%281%29%28-1%29%5E2+%2B+%28-1%29%5E2\"\r\n" );
document.write( "\"+0+=+-4A+%2B+1+%2B4C+%2B+1\"\r\n" );
document.write( "\"+0+=+-4A+%2B+4C+%2B+2\"\r\n" );
document.write( "\"4A-4C=2\"\r\n" );
document.write( "\"2A-2C=1\"\r\n" );
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document.write( "The next easiest number you haven't substituted is x=2\r\n" );
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document.write( "\"+4%2A4+=+4A%2A%281%29%283%29%5E2+%2B+%283%29%5E2+%2B4C%2A%283%29%281%29%5E2+%2B+%281%29%5E2\"\r\n" );
document.write( "\"+16+=+4A%2A%289%29+%2B+9+%2B4%2AC%2A%283%29+%2B+1\"\r\n" );
document.write( "\"+16+=+36A+%2B+10+%2B12C\"\r\n" );
document.write( "\"+6+=+36A+%2B+12C\"\r\n" );
document.write( "\"+1+=+6A+%2B+2C\"\r\n" );
document.write( "\"+6A+%2B+2C+=1\"\r\n" );
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document.write( "Solve the system of equations:\r\n" );
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document.write( "\"system%282A-2C=1%2C6A+%2B+2C+=1%29\"\r\n" );
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document.write( "Get \"A=1%2F4\", \"C=-1%2F4\"\r\n" );
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document.write( "That's how you do it. If you run out of numbers to substitute for x that\r\n" );
document.write( "cause terms to be 0, substitute some numbers that DON'T cause any terms\r\n" );
document.write( "to be 0, and solve a system of equations.  Any numbers will do.  I just\r\n" );
document.write( "picked the next easiest numbers x=0 and x=2. \r\n" );
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document.write( "Edwin
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