# Question Video: Finding the Partial Fraction Decomposition With a Repeated Factor Mathematics

Determine the partial fraction decomposition of β1/(π₯Β²(π₯ β 1)).

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### Video Transcript

Determine the partial fraction decomposition of negative one over π₯ squared times π₯ minus one.

When we split a fraction into its partial fractions, weβre essentially reversing the process of adding rational expressions. Weβre looking to express our function as the sum of two or more simpler ones. Now, weβre going to need to be really careful here. We have a repeated factor of π₯ in our denominator. And so the factor of π₯ needs to appear twice. It appears once on its own and once squared.

And so we write negative one over π₯ squared times π₯ minus one as π΄ over π₯ plus π΅ over π₯ squared plus πΆ over π₯ minus one, where π΄, π΅, and πΆ are real constants to be calculated. Before we can calculate these constants though, weβre going to make the expression on the right-hand side look a little more like that on the left.

Weβre going to add the three fractions by creating a common denominator. The common denominator is going to be π₯ squared times π₯ minus one. And weβll achieve this by multiplying the first fraction by π₯ times π₯ minus one. We multiply the numerator and denominator of our second fraction by π₯ minus one and of our third fraction by π₯ squared. And so the right-hand side becomes π΄π₯ π₯ minus one over π₯ squared times π₯ minus one plus π΅ times π₯ minus one over π₯ squared times π₯ minus one plus πΆπ₯ squared over π₯ squared times π₯ minus one.

And of course now that the denominators of our three fractions are equal, we can simply add the numerators. And the numerator becomes π΄π₯ times π₯ minus one plus π΅ times π₯ minus one plus πΆπ₯ squared. Now, of course this is still equal to negative one over π₯ squared times π₯ minus one. Letβs rewrite this slightly as negative one over π₯ squared times π₯ minus one.

And we now see that we have two fractions that are equal and their denominators themselves are equal. This must mean that their numerators must also be equal. So negative one must be equal to π΄π₯ times π₯ minus one plus π΅ times π₯ minus one plus πΆπ₯ squared. But how do we find the values of π΄, π΅, and πΆ?

Well, one method we have is to substitute the zeros of π₯ squared times π₯ minus one in. One of the zeros is π₯ equals zero. So letβs substitute this in and see what happens. The left-hand side is simply negative one. Then on the right-hand side, we get π΄ times zero plus π΅ times negative one plus πΆ times zero, which simplifies to negative π΅.

And we now see that the whole purpose of substituting in the zeros of π₯ squared times π₯ minus one was to form an equation purely in terms of one of our constants. We multiply through by negative one, and we see π΅ must be equal to one. The next zero we choose is one. When we let π₯ be equal to one, we get negative one equals π΄ times zero plus π΅ times zero plus πΆ times one squared. In other words, πΆ is equal to negative one. So we found the values of π΅ and πΆ.

There are a couple of methods we can use to find the value of π΄. We could distribute the parentheses and equate coefficients. Alternatively, once we know the values of π΅ and πΆ, we can actually choose any value for π₯. Letβs try π₯ is equal to two. Substituting π₯ equals two, and we get negative one equals π΄ times two times one plus π΅ times one plus πΆ times four. Well, the right-hand side simplifies to two π΄ plus π΅ plus four πΆ.

But we also know that π΅ is one and πΆ is negative one. So our equation simplifies to negative one equals two π΄ minus three. And we solve for π΄ by adding three to both sides and dividing through by two. And we find π΄ is equal to one. By finding the values of π΄, π΅, and πΆ, we have determined the partial fraction decomposition of negative one over π₯ squared times π₯ minus one. Itβs one over π₯ plus one over π₯ squared minus one over π₯ minus one.