M1410704Part2

M1410704Part2 - (Section 7.4: Partial Fractions) 7.27 PART...

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(Section 7.4: Partial Fractions) 7.27 PART E: REPEATED (OR POWERS OF) LINEAR FACTORS Example Find the PFD for x 2 x + 2 ( ) 3 . Solution Step 1: The expression is proper, because 2, the degree of N x ( ) , is less than 3, the degree of D x ( ) . Warning: Imagine that N x ( ) and D x ( ) have been written out in standard form before determining the degrees. Step 2: Factor the denominator over R . (Done!) Step 3: Determine the required PFD Form. The denominator consists of repeated linear factors, so the PFD Form is given by: x 2 x + 2 ( ) 3 = A x + 2 + B x + 2 ( ) 2 + C x + 2 ( ) 3 We must “run up to the power.” Step 4: Multiply both sides of the equation by the LCD, x + 2 ( ) 3 , to obtain the basic equation. x 2 = A x + 2 ( ) 2 + B x + 2 ( ) + C
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(Section 7.4: Partial Fractions) 7.28 Step 5: Solve the basic equation for the unknowns, A , B , and C . Plug in x = 2 : x 2 = A x + 2 ( ) 2 + B x + 2 ( ) + C 2 ( ) 2 = 0 + 0 + C C = 4 Updated basic equation; we now know C = 4 : x 2 = A x + 2 ( ) 2 + B x + 2 ( ) + 4 We can use the “Match Coefficients” Method, or we can plug in a couple of other real values for x , as follows: Plug in x = 0 , say: x 2 = A x + 2 ( ) 2 + B x + 2 ( ) + 4 0 ( ) 2 = A 0 + 2 ( ) 2 + B 0 + 2 ( ) + 4 0 = 4 A + 2 B + 4 Let's divide both sides by 2 and switch sides. 2 A + B + 2 = 0 2 A + B = 2 Plug in x = 1 , say: x 2 = A x + 2 ( ) 2 + B x + 2 ( ) + 4 1 ( ) 2 = A 1 + 2 ( ) 2 + B 1 + 2 ( ) + 4 1 = A + B + 4 3 = A + B A + B = 3
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(Section 7.4: Partial Fractions) 7.29 We must solve the system: 2 A + B = 2 A + B =
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This note was uploaded on 09/08/2011 for the course MATH 141 taught by Professor Staff during the Fall '11 term at Mesa CC.

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M1410704Part2 - (Section 7.4: Partial Fractions) 7.27 PART...

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