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Unformatted text preview: n decomposition for this part. 3x3 + 7 x  4 (x 2 + 2) 2 = Ax + B Cx + D
+
x 2 + 2 ( x 2 + 2 )2 Adding the two terms up gives, © 2007 Paul Dawkins 277 http://tutorial.math.lamar.edu/terms.aspx College Algebra 3x3 + 7 x  4 (x 2 + 2) 2 ( Ax + B ) ( x 2 + 2 ) + Cx + D = (x 2 + 2) 2 Now, set the numerators equal and we might as well go ahead and multiply the right side out and
collect up like terms while we’re at it. 3 x 3 + 7 x  4 = ( Ax + B ) ( x 2 + 2 ) + Cx + D 3 x 3 + 7 x  4 = Ax 3 + 2 Ax + Bx 2 + 2 B + Cx + D
3 x 3 + 7 x  4 = Ax 3 + Bx 2 + ( 2 A + C ) x + 2 B + D
Setting coefficients equal gives, A=3
B=0
2A + C = 7
2 B + D = 4 In this case we got A and B for free and don’t get excited about the fact that B = 0 . This is not a
problem and in fact when this happens the remaining work is often a little easier. So, plugging
the known values of A and B into the remaining two equations gives, 2 ( 3) + C = 7 Þ C =1 2 ( 0 ) + D = 4 Þ D = 4 The partial fraction decomposition is then, 3x3 + 7 x  4 (x 2 + 2) 2 = 3x...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.
 Spring '12
 MrVinh

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