Video: Partial Fraction Decomposition

Express (π‘₯Β² βˆ’ 3)/((π‘₯Β² + 2)(π‘₯ βˆ’ 1)) in partial fractions.

07:34

Video Transcript

Express π‘₯ squared minus three over π‘₯ squared plus two times π‘₯ minus one in partial fractions.

This means, we need to write it as the sum of two fractions, where the denominators of our two fractions are the factors of the original denominator. Our denominator is already factored. And that means we know that we’ll have one denominator of π‘₯ squared plus two and one denominator of π‘₯ minus one. Notice that in our first denominator, we have an π‘₯ squared term. That means our numerator will be in the form 𝐴π‘₯ plus 𝐡. In our second denominator, we have a power of one, which means our numerator can just be a constant 𝐢.

To solve for 𝐴, 𝐡, and 𝐢, we want to get rid of all these denominators. And that means we’ll multiply the left side of the equation by the original denominator, π‘₯ squared plus two times π‘₯ minus one. But if we multiply the left side by the original denominator, we need to multiply both terms on the right side by the original denominator, π‘₯ squared plus two times π‘₯ minus one times 𝐴π‘₯ plus 𝐡 over π‘₯ squared plus two. And yes, we need to multiply the original denominator by that 𝐢 term, π‘₯ squared plus two times π‘₯ minus one times 𝐢 over π‘₯ minus one.

At this stage, a lot of things will cancel out. π‘₯ squared plus two over π‘₯ squared plus two, π‘₯ minus one over π‘₯ minus one. And we have [π‘₯ squared] minus three. From there, π‘₯ squared plus two over π‘₯ squared plus two, π‘₯ minus one over π‘₯ minus one. Our new statement says π‘₯ squared minus three equals π‘₯ minus one times 𝐴π‘₯ plus 𝐡 plus π‘₯ squared plus two times 𝐢. Let’s go ahead and clear off this stage to make more room. But I wanted to show you this step so you knew where we got this statement from.

At this point, we want to go ahead and distribute and expand π‘₯ times 𝐴π‘₯ equals 𝐴π‘₯ squared. π‘₯ times 𝐡 equals 𝐡π‘₯. Negative one times 𝐴π‘₯ equals negative 𝐴π‘₯. And negative one times 𝐡 equals negative 𝐡. Then we have 𝐢 times π‘₯ squared equals 𝐢π‘₯ squared, and 𝐢 times two, plus two 𝐢. π‘₯ squared minus three equals 𝐴π‘₯ squared plus 𝐡π‘₯ minus 𝐴π‘₯ minus 𝐡 plus 𝐢π‘₯ squared plus two 𝐢. And just like any regular polynomials, we can combine like terms. 𝐴π‘₯ squared plus 𝐢π‘₯ squared are like terms. 𝐡π‘₯ and 𝐴π‘₯ are like terms. And negative 𝐡 and two 𝐢 are like terms.

Once we combine like terms, we can do some simplifying. 𝐴π‘₯ squared plus 𝐢π‘₯ squared equals 𝐴 plus 𝐢 times π‘₯ squared. 𝐡π‘₯ minus 𝐴π‘₯ equals 𝐡 minus 𝐴 times π‘₯. And our final term, two 𝐢 minus 𝐡 can’t be simplified any further. We wanna bring down this π‘₯ squared minus three. But I also want to add one term in the middle. We have zero π‘₯ to the first power. On the left side of our equation, we have one π‘₯ squared plus zero π‘₯ to the first power minus three. Because we know that the left side and the right side of these equations have to be equal to each other, we know that the coefficient of π‘₯ squared must be equal to one. And that means that 𝐴 plus 𝐢 equals one. It also means that 𝐡 minus 𝐴 has to be equal to zero, and that two 𝐢 minus 𝐡 equals negative three. 𝐴 plus 𝐢 equals one. 𝐡 minus 𝐴 equals zero. And two 𝐢 minus 𝐡 equals negative three.

In our middle equation, if I add 𝐴 to both sides, I see that 𝐡 must be equal to 𝐴. Our other two equations are in the term 𝐴𝐢 and then 𝐴𝐡. It would be helpful if we could exchange this 𝐡 in the third equation for an 𝐴. Because 𝐴 and 𝐡 are equal, we can substitute 𝐴 for 𝐡 and then say two 𝐢 minus 𝐴 equals negative three. In our first equation, if we subtract 𝐢 from both sides, we find that 𝐴 equals one minus 𝐢. I already know that two 𝐢 minus 𝐴 equals negative three. And I know that 𝐴 equals one minus 𝐢. Two 𝐢 minus one minus 𝐢 equals negative three. Two 𝐢 minus one plus 𝐢 equals negative three. Two 𝐢 plus 𝐢 equals three 𝐢.

Then we add one to both sides, negative three plus one equals negative two. We can say that three 𝐢 equals negative two, and then divide both sides by three. And we found that 𝐢 equals negative two-thirds. We know that 𝐴 equals one minus 𝐢. We plug in negative two-thirds for 𝐢. And we take the whole number one and write it as the fraction three over three. Three over three minus negative two over three equals three over three plus two over three. 𝐴 equals five-thirds, and 𝐡 equals 𝐴, which means 𝐡 also equals five-thirds. We’re gonna need to clear some more space.

Now that we found our 𝐴, 𝐡, and 𝐢 values, we can plug them back in to the original partial fraction equation we found. That would look like this. Five-thirds π‘₯ plus five-thirds over π‘₯ squared plus two plus negative two-thirds over π‘₯ minus one. Now, we usually don’t love leaving fractions in our numerator. Let’s consider the first fraction, five-thirds π‘₯ plus five-thirds, could be written as five π‘₯ plus five over three. If we divide that by π‘₯ plus two, that’s the same thing as multiplying it by one over π‘₯ squared plus two. Then we could say five π‘₯ plus five over three times π‘₯ squared plus two.

We can do the same thing for the other sum. Negative two-thirds divided by π‘₯ minus one equals negative two-thirds times one over π‘₯ minus one, which equals negative two over three times π‘₯ minus one. And then we can make one final adjustment. Instead of adding negative two over three times π‘₯ minus one, we can subtract two over three times π‘₯ minus one.

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