document.write( "Question 654936: Find the partial decomposition of (3x^3-5x^2-3x+15)/(x^3(x-5)) \n" ); document.write( "
Algebra.Com's Answer #408892 by Edwin McCravy(20060) $\"\"$ $\"About$
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document.write( "Since x³ has factors x³, x², and x we must include a \r\n" );
document.write( "fraction with each of these as a denominator:\r\n" );
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document.write( " =  +  +  + \r\n" );
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document.write( "Clear of fractions:\r\n" );
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document.write( "3x³ - 5x² - 3x + 15 = A(x-5) + Bx(x-5) + Cx²(x-5) + Dx³\r\n" );
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document.write( "Substitute x = 0 to eliminate the last three terms on the right:\r\n" );
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document.write( "3·0³ - 5·0² - 3·0 + 15 = A(0-5) + B·0(0-5) + C·0²(0-5) + D·0³\r\n" );
document.write( "                    15 = -5A\r\n" );
document.write( "                    -3 = A\r\n" );
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document.write( "Substitute that:\r\n" );
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document.write( "3x³ - 5x² - 3x + 15 = -3(x-5) + Bx(x-5) + Cx²(x-5) + Dx³\r\n" );
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document.write( "Substitute x = 5 to eliminate the first three terms on the right:\r\n" );
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document.write( "3·5³ - 5·5² - 3·5 + 15 = -3(5-5) + B·5(5-5) + C·5²(5-5) + D·5³\r\n" );
document.write( "3·125 - 5·25 - 15 + 15 = -3(0) + B·5(0) + C·5²(0) + D·125\r\n" );
document.write( "                   250 = 125D\r\n" );
document.write( "                     2 = D\r\n" );
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document.write( "Substitute that:\r\n" );
document.write( "3x³ - 5x² - 3x + 15 = -3(x-5) + Bx(x-5) + Cx²(x-5) + 2x³\r\n" );
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document.write( "Now that we have substituted all possible numbers that will eliminate \r\n" );
document.write( "terms, we now substitute any other convenient numbers which WON'T\r\n" );
document.write( "eliminate terms, but which will give us equations in the letters we\r\n" );
document.write( "haven't determined.  We haven't substituted 1, so let's choose x = 1: \r\n" );
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document.write( "Substitute x = 1\r\n" );
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document.write( "3·0³ - 5·0² - 3·0 + 15 = A(0-5) + B·0(0-5) + C·0²(0-5) + D·0³\r\n" );
document.write( "3·1³ - 5·1² - 3·1 + 15 = -3(1-5) + B·1(1-5) + C·1²(1-5) + 2·1³\r\n" );
document.write( "        3 - 5 - 3 + 15 = -3(-4) + B(-4) + C(1-5) + 2\r\n" );
document.write( "                    10 = 12 - 4B - 4C + 2\r\n" );
document.write( "                    10 = 14 - 4B - 4C  \r\n" );
document.write( "               4B + 4C = 4\r\n" );
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document.write( "Divide through by 4\r\n" );
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document.write( "                 B + C = 1\r\n" );
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document.write( "We haven't substituted 2, so let's substitute x = 2:\r\n" );
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document.write( "3·2³ - 5·2² - 3·2 + 15 = -3(2-5) + B·2(2-5) + C·2²(2-5) + 2·2³\r\n" );
document.write( "    3·8 - 5·4 - 6 + 15 = -3(-3) + B·2(-3) + C·4(-3) + 2·8\r\n" );
document.write( "      24 - 20 - 6 + 15 = 9 - 6B - 12C + 16\r\n" );
document.write( "                    13 = 25 - 6B - 12C\r\n" );
document.write( "              6B + 12C = 12\r\n" );
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document.write( "Divide through by 6\r\n" );
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document.write( "                B + 2C = 2\r\n" );
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document.write( "Solve the resulting system of equations by substitution or elimination\r\n" );
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document.write( "                B +  C = 1\r\n" );
document.write( "                B + 2C = 2\r\n" );
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document.write( "Get B = 0, C = 1\r\n" );
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document.write( "So we have A = -3, B = 0, C = 1, D = 2\r\n" );
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document.write( "Answer:\r\n" );
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document.write( " =  +  +  +  \r\n" );
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document.write( " =  +  +  + \r\n" );
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document.write( "Omitting the 0 term and bringing the negative sign out front of the\r\n" );
document.write( "first term:\r\n" );
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document.write( " = — +  + \r\n" );
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document.write( "Edwin
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