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document.write( "First we factor the denominator:\r\n" );
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document.write( "x³ + 5x² + 7x + 3\r\n" );
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document.write( "if we can find a zero of that polynomial then (x - that zero) is\r\n" );
document.write( "a factor.\r\n" );
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document.write( "Since the the leading coefficient (the coefficient of x³) is 1,\r\n" );
document.write( "all possible rational zeros are ± factor of the constant term 3\r\n" );
document.write( "So they are ±1, ±3\r\n" );
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document.write( "We try 1 using synthetic division\r\n" );
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document.write( "1 | 1 5 7 3\r\n" );
document.write( " | 1 6 13\r\n" );
document.write( " 1 6 13 16\r\n" );
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document.write( "So 1 is not a zero.\r\n" );
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document.write( "We try -1 using synthetic division\r\n" );
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document.write( "-1 | 1 5 7 3\r\n" );
document.write( " | -1 -4 -3\r\n" );
document.write( " 1 4 3 0\r\n" );
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document.write( "Yes, that left a zero remainder, So (x + 1)\r\n" );
document.write( "is a factor and so is (x² + 4x + 3) from the\r\n" );
document.write( "bottom line of the synthetic division. So now\r\n" );
document.write( "we have factored \r\n" );
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document.write( "x³ + 5x² + 7x + 3\r\n" );
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document.write( "as\r\n" );
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document.write( "(x + 1)(x² + 4x + 3)\r\n" );
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document.write( "And we can further factor the quadratic in the\r\n" );
document.write( "second parentheses:\r\n" );
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document.write( "(x + 1)(x + 1)(x + 3)\r\n" );
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document.write( "or\r\n" );
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document.write( "(x + 1)²(x + 3)\r\n" );
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document.write( "Now the original problem:\r\n" );
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document.write( "becomes 
\r\n" );
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document.write( " = 
+ 
+ 
\r\n" );
document.write( "\r\n" );
document.write( "(Notice that when you have a factor to a power larger than 1,\r\n" );
document.write( "you must have a fraction with that power and all lower powers.)\r\n" );
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document.write( "Multiply through by the LCD = (x + 1)²(x + 3)\r\n" );
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document.write( "4x² = A(x+1)(x+3) + B(x+3) + C(x+1)² \r\n" );
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document.write( "Multiplying all that out gives:\r\n" );
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document.write( "4x² = Ax² + 4Ax + 3A + Bx + 3B + Cx² + 2Cx + C\r\n" );
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document.write( "Equating the x² terms on each side:\r\n" );
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document.write( "4x² = Ax² + Cx²\r\n" );
document.write( " 4 = A + C\r\n" );
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document.write( "Equating the x terms on each side (there are none on the left so we put 0):\r\n" );
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document.write( "0 = 4Ax + Bx + 2Cx \r\n" );
document.write( "0 = 4A + B + 2C\r\n" );
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document.write( "Equating the constant terms on each side: (none on the left so we put 0)\r\n" );
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document.write( "0 = 3A + 3B + C\r\n" );
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document.write( "So we have this system of equations:\r\n" );
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document.write( "
\r\n" );
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document.write( "Solve that system of equations and get A=-5, B = 2, C = 9\r\n" );
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document.write( "So the partial fraction decomposition is:\r\n" );
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document.write( " = 
+ 
+ 
\r\n" );
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document.write( "-------------------------------------------------- \r\n" );
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document.write( "There is an easier way to find A, B and C.\r\n" );
document.write( "It's the substitution method. When you get here:\r\n" );
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document.write( "4x² = A(x+1)(x+3) + B(x+3) + C(x+1)²\r\n" );
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document.write( "Don't multiply it out.\r\n" );
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document.write( "Substitute x = -1 to make the 1st and 3rd terms on the right 0\r\n" );
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document.write( "4(-1)² = A(-1+1)(-1+3) + B(-1+3) + C(-1+1)²\r\n" );
document.write( " 4 = 0 + 2B + 0\r\n" );
document.write( " 4 = 2B\r\n" );
document.write( " 2 = B\r\n" );
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document.write( "Substitute x = -3 to make the 1st and 2nd terms on the right 0\r\n" );
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document.write( "4(-3)² = A(-3+1)(-3+3) + B(-3+3) + C(-3+1)²\r\n" );
document.write( " 36 = 0 + 0 + 4C\r\n" );
document.write( " 36 = 4C\r\n" );
document.write( " 9 = C\r\n" );
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document.write( "Substitute another easy value, say x = 0, and B=2 and C=9\r\n" );
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document.write( "4(0)² = A(0+1)(0+3) + 2(0+3) + 9(0+1)²\r\n" );
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document.write( " 0 = 3A + 6 + 9\r\n" );
document.write( " 0 = 3A + 15\r\n" );
document.write( " -15 = 3A\r\n" );
document.write( " -5 = A\r\n" );
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document.write( " = + + \r\n" );
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document.write( "With this method you don't have to multiply the whole \r\n" );
document.write( "thing out or solve a system of equattions. This way is\r\n" );
document.write( "much easier than equating like terms and solving a \r\n" );
document.write( "system of equations. But some teachers won't let you use \r\n" );
document.write( "this method, even though it always works. Some won't \r\n" );
document.write( "allow it because the original problem is undefined for \r\n" );
document.write( "the values you have to substitute to make those \r\n" );
document.write( "expressions = 0, for it makes denominators = 0 in the\r\n" );
document.write( "original problem.\r\n" );
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document.write( "Edwin
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document.write( "![\"\"](\"/images/stars/stars-13.gif\")
