Partial Fraction Decomposition Find the partial fraction decomposition of the rational function.

Partial Fraction Decomposition Find the partial fraction...

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to express a complicated rational function as a sum of simpler fractions. This is especially useful in integration, differential equations, and Laplace transforms because it breaks down complex expressions into terms that are easier to work with.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. Understanding their structure is important because it allows you to decompose them into partial fractions when the degree of the numerator is less than that of the denominator, making them simpler to integrate or analyze.

Repeated Linear Factors

When the denominator of a rational function includes repeated linear factors, each occurrence of the factor contributes a separate term in the partial fraction expansion. For a factor raised to a power, such as (x - a)^n, the decomposition must include fractions with denominators (x - a), (x - a)^2, ..., up to (x - a)^n.

Setting Up Equations for Coefficients

The process of partial fraction decomposition involves expressing the original function as a sum of unknown coefficients over simpler denominators, then clearing the denominators and equating coefficients of like powers of x. This method results in a system of linear equations that, once solved, gives the values of the unknown coefficients in the decomposition.

Transcript

00:01 In this problem, we're being asked to find the partial fraction decomposition of the given fraction.

00:05 Well, if you notice, our denominator is already in factored form, which is great.

00:10 So let's start with our first factor, which is a linear factor of x minus 1, and it's a distinct linear factor.

00:17 However, it's getting repeated three times.

00:20 So to write this in its partial fractions, we're going to have a over x minus 1 plus b over the quantity of x minus 1 squared, plus c over the quantity of x minus 1 raised to the third power.

00:37 Okay, so now we can move to our next factor, which is the distinct linear factor x plus 2.

00:42 So for this, we'll just need plus d all over x plus 2.

00:48 And remember, we know this is all equal to our original fraction, negative 10x squared plus 27x minus 14, all over the quantity of x minus 1 to the third times x plus 2.

01:03 Now, our next step is we're going to multiply both sides of our equation by our common denominator, which in this case is x minus 1 to the 3rd times x plus 2.

01:15 And same thing on the other side.

01:17 We're going to have x minus 1 to the 3rd times x plus 2.

01:22 So now, remember, the reason why we're doing this is because on the right hand side of the equation, this x minus 1 to the 3rd factor will cancel, as well as the x plus 2 factor will cancel.

01:31 So now we won't have the fraction.

01:34 Now let's start by looking at the left -hand side.

01:37 Well, our first fraction, a over x minus 1, is missing an x -1 to the second power and an x plus 2.

01:44 So we're going to have a times x -minus 1 to the second times x plus 2.

01:51 Now, our second fraction, b over the quantity of x -minus 1 squared, is missing an x -minus 1 factor and an x -plus -2 factor.

01:59 So we're going to have b times x -1 times x -plus -2.

02:06 Now for our third fraction, the c over x minus 1 to the third, it's missing the x plus 2 factor.

02:13 So we're going to have plus c times x plus 2.

02:18 And for our last fraction, d over x plus 2, it's missing the x minus 1 to the third power.

02:24 So we're going to have d times x minus 1 to the third.

02:28 And remember, it's all equal to our numerator, negative 10 x squared plus 27x minus 14.

02:37 Well, what we need to start doing is we need to start simplifying this.

02:41 Well, as you can see, i've kind of ran out of room here.

02:45 So i'm going to start the new page.

02:47 And at the same time, i'm going to expand each of our multiplication.

02:51 So i'm going to start with this first a term.

02:53 It's getting multiplied by x minus 1 to the second times x plus 2.

02:58 So when i multiply those out, i'm now going to have a times x to the 3rd minus 3x.

03:07 Plus 2.

03:10 So now if we go back, i'm now going to multiply x minus 1 times x plus 2, and i'm going to bring down that b term.

03:18 So we're going to have plus b times, and when i multiply those two binomials, i get x squared plus x minus 2.

03:31 Now let's look at our next term, which is c times x plus 2.

03:35 Well, when we distribute that, we just get cx plus 2c.

03:38 So we have plus cx plus 2c.

03:44 Then for our next term, we have the quantity of x minus 1 raised to the third power.

03:51 So i'm going to expand that binomial.

03:53 So when i do that, i'm going to have plus d times x to the third minus 3x squared plus 3x minus 1.

04:02 And again, this is all still equal to, if we go back just to check, negative 10x squared plus 27x minus 14.

04:11 So negative 10x squared.

04:15 Plus 27x minus 14.

04:21 All right.

04:21 Well, now i'm just going to distribute.

04:23 So that way we won't have any of those parentheses left.

04:25 So when i do that, i'm going to have a x to the third minus 3ax plus 2a.

04:32 And then i'm going to have plus bx squared plus bx minus 2b.

04:40 And i'm just going to bring down the plus cx plus 2c.

04:44 And then i'll distribute the d.

04:45 So i'll get dx to the third, minus.

04:49 3dx squared plus 3dx minus d and it's all equal to negative 10x squared plus 27x minus 14.

05:03 All right.

05:04 So now what we're going to do is we're going to combine our light terms.

05:08 Well, let's start with our first term, ax to the third.

05:11 Well, it's only like term is plus dx to the third.

05:14 So i'm going to have ax to the third plus dx to the third.

05:19 Then i'm just going to go in order.

05:21 So i'm going to then do my x squared terms.

05:23 Well, i have a positive bx squared and a negative 3 dx squared.

05:28 So i can combine those.

05:29 So i'll have plus bx squared minus 3 dx squared.

05:35 All right.

05:36 Now for my x terms...

Please give Ace some feedback