Video: Partial Fraction Decomposition

The expression (3π‘₯ βˆ’ 2)/((π‘₯Β² + 4)(π‘₯ βˆ’ 3)) can be written in the form (𝐴π‘₯ + 𝐡)/(π‘₯Β² + 4) + 𝐢/(π‘₯ βˆ’ 3). Find the values of 𝐴, 𝐡, and 𝐢.

06:25

Video Transcript

The expression three π‘₯ minus two over π‘₯ squared plus four multiplied by π‘₯ minus three can be written in the form 𝐴π‘₯ plus 𝐡 over π‘₯ squared plus four plus 𝐢 over π‘₯ minus three. Find the values of 𝐴, 𝐡, and 𝐢.

What we’ve been asked to find here is the partial fractions decomposition of this expression. We’re taking it from a single expression into the sum of two fractions whose denominators are the factors of the original denominator, π‘₯ squared plus four and π‘₯ minus three. To do this, let’s first consider how we would go the other way. So how would we combine the two fractions on the right to a single fraction with a common denominator?

Well, for the first fraction, we need to multiply its denominator by π‘₯ minus three. And so, in order for this to be equivalent, we need to do the same in the numerator. For the second fraction, we’d need to multiply its denominator by π‘₯ squared plus four. So again, we’d have to do the same in the numerator. Doing so would give the right-hand side as 𝐴π‘₯ plus 𝐡 multiplied by π‘₯ minus three over π‘₯ squared plus four multiplied by π‘₯ minus three plus 𝐢 multiplied by π‘₯ squared plus four over π‘₯ minus three multiplied by π‘₯ squared plus four. And now, we have a common denominator for these two terms. So we could write them as a single fraction.

Now, the key point is that the fraction on the left and the fraction on the right now have exactly the same denominators. So if these two fractions are equal to one another, it must be the case that the numerators are equal to one another. So we can rewrite our equation using the numerators only, and we have a simplified equation. Three π‘₯ minus two equals 𝐴π‘₯ plus 𝐡 multiplied by π‘₯ minus three plus 𝐢 multiplied by π‘₯ squared plus four.

Now, we’re looking to find the values of 𝐴, 𝐡, and 𝐢. And to do this, we remember that this equation is true for all values of π‘₯. So the first approach we can try is to choose values of π‘₯ which will eliminate some of these constants. If we choose, for example, π‘₯ equals three, then π‘₯ minus three will be equal to zero. Which means we’ll be able to eliminate the first part of the expression on the right-hand side. On the left-hand side, we’ll have three multiplied by three minus two. And on the right-hand side, we’ll be left with 𝐢 multiplied by three squared plus four. This gives nine minus two equals 𝐢 multiplied by nine plus four or seven equals 13𝐢. And dividing through by 13, we see that 𝐢 is equal to seven over 13.

So we’ve found one of our three values. We then consider whether it’ll be possible to choose an π‘₯-value that makes the second term on the right-hand side equal to zero. Well, in fact, this isn’t possible because it would be required that π‘₯ squared is equal to negative four. And this isn’t the case for any real values of π‘₯. So we won’t be able to take this approach. What we can do instead then is consider a method called equating coefficients. We look at the coefficients of, for example, π‘₯ squared, π‘₯, and the constant term on the two sides of our equation. Before we can do this, though, we’ll need to distribute the parentheses on the right-hand side.

Doing so gives 𝐴π‘₯ multiplied by π‘₯ which is 𝐴π‘₯ squared, 𝐴π‘₯ multiplied by negative three which is negative three 𝐴π‘₯, 𝐡 multiplied by π‘₯ which is 𝐡π‘₯, and 𝐡 multiplied by negative three which is negative three 𝐡. In the second set, we have 𝐢 multiplied by π‘₯ squared which is 𝐢π‘₯ squared and 𝐢 multiplied by four which is four 𝐢. Let’s now consider the coefficients of π‘₯ squared first of all. On the left-hand side, there are no π‘₯ squareds. So the coefficient is zero. And on the right-hand side, we have 𝐴 multiplied by π‘₯ squared plus 𝐢 multiplied by π‘₯ squared. So the coefficient of π‘₯ squared on the right-hand side is 𝐴 plus 𝐢.

Now, we know the value of 𝐢. We’ve already found it’s seven over 13. So we can substitute this value of 𝐢 in order to give an equation we can solve for 𝐴. We have zero equals 𝐴 plus seven over 13. And subtracting seven over 13 from each side, we find that the value of 𝐴 is negative seven over 13. To find the value of 𝐡, we can consider either the coefficient of π‘₯ or the value of the constant term because we have both 𝐡π‘₯ and negative three 𝐡 on the right-hand side. And the choice is somewhat arbitrary. Let’s consider then the constant term. On the left-hand side, the constant term is negative two. On the right-hand side, we have negative three 𝐡 and four 𝐢, both of which are independent of π‘₯. So our constant term is negative three 𝐡 plus four 𝐢.

To find the value of 𝐡, we rearrange this equation. And in doing so, we find that 𝐡 is equal to four 𝐢 plus two all over three. Remember, we’ve already found the value 𝐢. It’s seven over 13. So substituting, we have that 𝐡 is equal to four multiplied by seven over 13 plus two all over three. Four multiplied by seven over 13 is 28 over 13. And then, we can think of the integer two as 26 over 13. We therefore have 28 over 13 plus 26 over 13 all over three, which is equivalent to 28 plus 26 over 13 multiplied by three. That gives 54 over 39 which simplifies to 18 over 13 by dividing both the numerator and denominator by three.

So we’ve found the values of all three constants 𝐴, 𝐡, and 𝐢, both by choosing values of π‘₯ which will eliminate certain parts of our expression and by comparing the coefficients of π‘₯ squared and the constant term. It would be a useful check of our work to make sure that the values we’ve found for 𝐴, 𝐡, and 𝐢 do indeed work when comparing the final terms. So that would be the π‘₯ terms. We’ve found that the value of 𝐴 is negative seven over 13. The value of 𝐡 is 18 over 13. And the value of 𝐢 is seven over 13.

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