Find the partial fraction decomposition of each rational expression. (x^2)/((x-1)^2(x+1))

Name: Problem 19, Chapter 8: College Algebra
Uploaded: 2020-06-20T19:40:45-08:00
Duration: 6 min 24 s
Channel: Jason Schachner
Description: Find the partial fraction decomposition of each rational expression. (x^2)/((x-1)^2(x+1))

Find the partial fraction decomposition of each rational...

Key Concepts

Coefficient Comparison Method

The coefficient comparison method is a technique used to determine unknown constants in the decomposed form by equating the coefficients of corresponding powers of the variable on both sides of an identity. This systematic approach is essential in verifying and finding the precise form of the decomposition.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a complex rational expression as a sum of simpler fractions, each with simpler denominators. This is particularly useful in calculus and algebra for integrating functions or simplifying expressions.

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. Mastering their manipulation, especially through techniques like partial fraction decomposition, is crucial for solving equations and performing integrations in higher mathematics.

Repeated Linear Factors

Repeated linear factors occur when a factor in the denominator is raised to a power greater than one. In partial fraction decomposition, each repeated factor necessitates a series of terms corresponding to its different powers, ensuring the complete decomposition of the expression.

Transcript

00:01 In order to obtain our partial fraction decomposition, we must first look at the denominator of our rational expression.

00:10 Now, we have x minus 1 squared as a repeating function, x plus 1 is a non -repeating function.

00:16 So we can start writing our partial fraction decomposition as a over.

00:23 So let's look at x minus 1 square that's repeated.

00:26 So we'll write x minus 1 plus b over.

00:32 X minus 1 squared plus now look at our non -repeating x plus 1 function so plus some number c over x plus 1 now let's multiply everything by our denominator of x minus 1 squared times x plus 1 so we're going to get x squared is equal to a times x minus 1 times x plus 1 plus b times x plus 1 plus c times x minus 1 squared.

01:19 Now let's multiply everything out on the right side.

01:23 So the next step we're going to write x squared is equal to.

01:26 Now x minus 1 times x plus 1 is x squared minus 1 times x squared minus 1 times a.

01:31 We get a x squared minus a plus b times x plus 1 is going to be b x plus b plus b plus now x minus 1 squared is x squared minus 2x plus 1 times c we have plus c x squared minus 2 cx plus c plus c now we're going to group our terms to yield.

02:05 So we're going to look at our x squared.

02:10 So we're going to get a plus c x squared plus now let's look at our x is here so plus b minus 2c x.

02:28 Now what's left minus a plus b plus b plus c x.

02:28 Now what's left minus a plus b plus c.

02:35 So now when we equate the coefficients on both sides of the equal sign, we're going to get this.

02:43 So we're going to look at here.

02:44 So what's in front of x squared on both sides? so here we could just say one x squared.

02:49 So a plus c is going to equal one.

02:55 There is no x and there are no other rational numbers.

03:00 So we could say that b minus 2c is equal to zero and that minus a plus b plus c is going to to equal zero.

03:13 Now we need to solve a system of linear equations.

03:16 Let's look at b minus 2c equals 0 and minus a plus b equals 0...

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