Question

Find the partial fraction decomposition of \frac{z+1}{z^3 - z^2 - 6z}.

          Find the partial fraction decomposition of \frac{z+1}{z^3 - z^2 - 6z}.
        
Find the partial fraction decomposition of (z+1)/(z^3 - z^2 - 6z).

Added by Joshua L.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Find the partial fraction decomposition of (z+1)/(z^(3)-z^(2)-6z) Find the partial fraction decomposition of z+1 z3 - z2 -6z
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Transcript

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00:04 We're asked to find the partial fraction decomposition of this crazy looking thing.
00:10 We want to find the individual fractions that we added together to get this combined monstrosity here.
00:15 So the first thing i want to do is i want to factor out that denominator.
00:18 So let me find its original factors.
00:20 I can see right away i can factor out an x, in which case i'll be left with x squared minus 4x plus 3.
00:27 And that trinomial can also be factored into the product of two linear pieces.
00:32 It would be x minus three times x minus 1.
00:36 All right, so there's my denominator factored.
00:40 So what i'm going to do is i'm going to take my original here, 3x squared minus 7x plus 6 divided by this denominator, x times x minus 3 times x minus 1.
00:51 And i'm going to say i know that i can set it equal to a over x.
00:56 So that's some constant over this linear term here, x, plus, and now b was over the x minus 3 and then plus c over the x minus 1 and of course these can be interchanged maybe you had a over x minus 3 b over x and c over x minus 1 doesn't matter the order that we start them off in all right so now what do we need to do well i'm going to multiply everything through by my common denominator so i'm going to multiply everything by x times x minus 3 times x minus 1 and for this first side here, i'm just going to be left with my numerator, 3x squared minus 7x plus 6, because everything in the top cancels out everything in the bottom.
01:40 On this side, when i multiply it by my first term here, a over x, my x is cancel, but a still has to be multiplied by x minus 3 times x minus 1.
01:51 For the b term, the x minus 3 cancels, so he still needs to be multiplied by x and x minus 1.
01:59 And then my c term has to be multiplied by x and x minus 3.
02:04 All right.
02:05 So there's our equation that we're going to need to solve here.
02:08 Now the nice thing is we want this to work for any value of x.
02:12 So i'm going to pick some convenient values of x.
02:15 So for example, i'm going to let x equals 0.
02:18 And i'll show you what i mean by convenient values.
02:21 When i substitute in a 0, i get 0 minus 0, which is 0 plus 6.
02:25 I just get the simplified version.
02:27 6 is equal to.
02:28 0 minus 3 is negative 3.
02:31 0 minus 1 is negative 1.
02:33 Negative 3 times negative 1 is positive 3.
02:35 So i get 3a.
02:37 And when i let x equal 0, this whole term drops out and this whole term drops out.
02:43 So there's my simplified equation here, plus 0 plus 0.
02:47 So i just get 3a is equal to 6, which means that a is equal to 2...
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