Lesson Video: Partial Fractions: Nonrepeated Linear Factors Mathematics

Video Transcript

Partial Fractions: Nonrepeated Linear Factors

In this video, we’re going to discuss partial fraction decomposition. We’re going to discuss what we mean by this. And we’re going to focus on the case where our denominator factors to give us nonrepeated linear factors. We’re going to go through the entire method step by step. And we’ll discuss what happens when the degree of the polynomial in our numerator is greater than or equal to the degree of the polynomial in our denominator. Before we start with partial fraction decomposition, let’s remember how to add two rational functions together.

There’s a few different ways of doing this. For example, we could use cross multiplication. We multiply the denominator of our first fraction and the numerator of our second fraction. Then we add onto this the numerator of our first fraction multiplied by the denominator of our second fraction. This would then give us our new numerator. Finally, we would then multiply our two denominators together. This would give us our new denominator. And this is one way of adding these two fractions together. However, we’re going to focus on a different method. To add two fractions together, we want the denominators of both fractions to be equal. Then we just need to add the numerators. To do this, we want to find the lowest common multiple of our denominators. This can often be very tricky. However, we have two unique linear factors.

So in this case, we just need to multiply our denominators together. So how does this help us add these two fractions together? We want the denominator of our second fraction to be 𝑥 minus one times 𝑥 plus two. So we’ll multiply the entire second fraction by 𝑥 plus two divided by 𝑥 plus two. Of course, this is the same as multiplying it by one. But this will allow us to add our two fractions together. And we also want our first fraction to have the denominator 𝑥 minus one times 𝑥 plus two. So we’ll multiply our numerator and our denominator through by 𝑥 minus one. We can now simplify this expression. Let’s start with our first fraction. We’ll just multiply the numerator and the denominator by 𝑥 minus one. This gives us three times 𝑥 minus one divided by 𝑥 minus one times 𝑥 plus two.

We can do the same with our second fraction. We multiply the numerator and the denominator by 𝑥 plus two. This gives us one times 𝑥 plus two divided by 𝑥 minus one times 𝑥 plus two. And now we can see we’ve made the denominators of our two fractions equal. So we can just add our numerators together. Doing this gives us the following expression. And we can then simplify our numerator. In the second term in our numerator, multiplying by one does not change its value.

Next, in the first term in our numerator, we can distribute three over our parentheses. We see this will give us three 𝑥 minus three. So we’ve now rewritten our fraction as three 𝑥 minus three plus 𝑥 plus two all divided by 𝑥 minus one times 𝑥 plus two. And we can then simplify this further. We have three 𝑥 plus 𝑥 is equal to four 𝑥 and negative three plus two is equal to negative one. So this gives us four 𝑥 minus one divided by 𝑥 minus one times 𝑥 plus two.

So let’s think about what we’ve done. We’ve added our two fractions together into one new fraction. Partial fraction decomposition will be a lot like the reverse of this process. We’ll start with one rational function. We’ll want to factor our denominator. Then we want to reverse this process to write it as the sum of two fractions. In particular, we’ll be interested in making sure that our numerators are constants.

And there’s one thing worth highlighting about this point. If our numerator is a constant, then let’s think about what happens to our numerators. We multiply this by the lowest common multiple of the denominators. We can then ask the question, what happens to the degree of the polynomial in our numerator compared to the degree of the polynomial in our denominator?

Well, let’s consider where the terms in our numerators came from. They’ll be the constants in our numerators. And remember, we wanted to make the denominator of each fraction the lowest common multiples of our denominator. So we multiplied each of our numerators by the remaining part of this lowest common denominator. So our numerator is made up of polynomials with smaller degree than our denominator. So if we try and reverse this process in this way, we must have our numerator is a smaller degree polynomial than our denominator. Otherwise, this won’t work. Let’s now look at some examples.

The expression two 𝑥 plus one divided by 𝑥 plus two times 𝑥 plus three can be written in the form 𝐴 divided by 𝑥 plus three plus 𝐵 divided by 𝑥 plus two. Find the values of 𝐴 and 𝐵.

We’re given an expression which is a linear function divided by a quadratic function, and this quadratic function is fully factored. We’re told we can write this in the form 𝐴 divided by 𝑥 plus three plus 𝐵 divided by 𝑥 plus two, where 𝐴 and 𝐵 are some constants. We need to find the values of 𝐴 and 𝐵. First, the question tells us we can rewrite this expression in this form, so they must be equal. We just need to find the values of 𝐴 and 𝐵 which make this expression true.

So let’s start by adding these two expressions together. We can see their denominators are not equal, so we’ll need to make them equal so we can add these together. To do this, we’ll multiply our first fraction by 𝑥 plus two divided by 𝑥 plus two. This way, the denominator of our first fraction will be 𝑥 plus three times 𝑥 plus two, which is worth noting is the same as the expression we have on the left. And we want to do the same with our second fraction. We’ll see we’ll need to multiply this by 𝑥 plus three divided by 𝑥 plus three. This gives us the following expression.

Now that our denominators are equal, we can just add the numerators together. This gives us 𝐴 times 𝑥 plus two plus 𝐵 multiplied by 𝑥 plus three all divided by 𝑥 plus two times 𝑥 plus three. But now we can see something. The denominator we have on the left-hand side of our expression and the denominator we have on the right-hand side of our expression are equal. So for these two functions to be equal, the numerators must also be equal. In fact, we have something even stronger. These must be equal for all values of 𝑥. So equating the two numerators, we get two 𝑥 plus one is equal to 𝐴 times 𝑥 plus two plus 𝐵 multiplied by 𝑥 plus three.

It’s worth pointing out here sometimes you’ll see this written as an equivalence relation. This is to reiterate the fact that this must be true for all values of 𝑥. However, we’ll just write this as equals and keep this fact in mind. We now want to solve for our values of 𝐴 and 𝐵. There are two main methods for doing this. We could equate the coefficients on the left- and right-hand side of our equation. But it’s often easier to substitute values of 𝑥 into this expression. For example, we can see 𝐴 is multiplied by 𝑥 plus two. So we could eliminate 𝐴 by multiplying it by zero. In other words, if we substitute 𝑥 is equal to negative two into this expression, we then multiply 𝐴 by zero. We eliminate it from our equation.

In fact, we can see we can do the same thing for 𝐵. We substitute 𝑥 is equal to negative three. Let’s start by substituting 𝑥 is equal to negative two. We get two times negative two plus one is equal to 𝐴 multiplied by negative two plus two plus 𝐵 times negative two plus three. We can then simplify this expression. First, two times negative two plus one is equal to negative three. Of course, our coefficient of 𝐴 simplifies to give us zero and our coefficient of 𝐵 simplifies to give us one. So, in fact, this entire equation simplifies to give us negative three is equal to 𝐵.

We’ll now do the same and eliminate our variable 𝐵. We’ll substitute 𝑥 is equal to negative three. This gives us two times negative three plus one is equal to 𝐴 multiplied by negative three plus two plus 𝐵 times negative three plus three. If we evaluate this expression, this time, it gives us negative five is equal to negative 𝐴. We can then multiply both sides of this equation through by negative one. This gives us that 𝐴 is equal to five. Therefore, we were able to show for the expression given to us in the question to be written in the form 𝐴 over 𝑥 plus three plus 𝐵 over 𝑥 plus two, we must have 𝐴 is equal to five and 𝐵 is equal to negative three.

In this question, we were able to take our expression and write it in a new form by using partial fraction decomposition. However, we were given the form we need to write it in. We now need to find out how we would approach this problem if we’re not given the form to write it in.

Express one over 𝑥 squared minus one in partial fractions.

The question wants us to express one over 𝑥 squared minus one in partial fractions. There are two things we always need to do when we’re asked to use partial fractions. The first thing we need to do is check the degrees of the polynomial in our numerator and our denominator, since we can only use partial fractions if the degree of the polynomial in our numerator is smaller than the one in the denominator. In this case, the numerator is a constant. So its highest power of 𝑥 is 𝑥 to the zeroth power. So we say its degree is zero. And in our denominator, the term with the highest power of 𝑥 is 𝑥 squared. So we say its degree is two. So we’ve done our first check.

The second thing we need to do is fully factor our denominator. In our case, our denominator is a difference between squares, so we can factor this as 𝑥 minus one times 𝑥 plus one. We’re now ready to start using partial fraction decomposition. First, we’ll write our expression as one divided by 𝑥 minus one times 𝑥 plus one. In this case, we can see we have two unique factors in our denominator. So we’ll try and write this in the form 𝐴 divided by 𝑥 minus one plus 𝐵 divided by 𝑥 plus one, where 𝐴 and 𝐵 are constants. We want to try and find our values of 𝐴 and 𝐵, so we’re going to need to add these two fractions together.

To add these two factions together, their denominators need to be equal. So we’ll multiply our first fraction by 𝑥 plus one divided by 𝑥 plus one and our second fraction by 𝑥 minus one divided by 𝑥 minus one. This gives us the following expression. Now that their denominators are equal, to add these two terms together, we just need to add the numerators. Adding the numerators of these two terms together, we get 𝐴 times 𝑥 plus one plus 𝐵 multiplied by 𝑥 minus one divided by 𝑥 minus one times 𝑥 plus one. But now we can see the denominator of the expression on the left-hand side of our equation and the denominator of the expression on the right-hand side of our equation are equal.

So the only way for these two functions to be equal is if their numerators are equal. So let’s equate the numerators on both sides of the equation. This gives us one is equal to 𝐴 times 𝑥 plus one plus 𝐵 times 𝑥 minus one. We could find our values of 𝐴 and 𝐵 by equating coefficients. However, remember, this must be true for any value of 𝑥. In particular, we could substitute 𝑥 is equal to negative one. Then, our coefficient of 𝐴 becomes zero. So we eliminate it from the equation. Substituting 𝑥 is equal to negative one, we get one is equal to 𝐴 times negative one plus one plus 𝐵 multiplied by negative one minus one.

Of course, as we said earlier, our coefficient of 𝐴 simplifies to give us zero. So this simplifies to give us that one is equal to negative two 𝐵. And we can divide this equation through by negative two to get that 𝐵 is equal to negative one-half. We can do something similar to eliminate our variable 𝐵. We substitute 𝑥 is equal to one. This will make the coefficient of 𝐵 equal to zero. Substituting 𝑥 is equal to one, we get one is equal to 𝐴 times one plus one plus 𝐵 multiplied by one minus one. This time, our coefficient of 𝐵 simplifies to give us zero. So we can simplify this equation to get one is equal to two 𝐴. And we divide this equation through by two to see that 𝐴 is equal to one-half.

The last thing we’ll do is substitute our value of 𝐴 is equal to one-half and our value of 𝐵 is equal to negative one-half into our expression in partial fractions. So we want to substitute 𝐴 is equal to one-half and 𝐵 is equal to negative one-half into this expression. Let’s start with 𝐴 is equal to one-half. However, we could simplify this expression. Instead of having one-half in our numerator, we’ll have one in our numerator and two in our denominator.

We now want 𝐵 is equal to negative one-half. And we can do something similar. Instead of having one-half in our numerator, we’ll have one in our numerator and two in our denominator. However, since our numerator is negative, we’re going to need to subtract this value, and this is our final answer. Therefore, by using partial fractions, we were able to express one over 𝑥 squared minus one as one over two times 𝑥 minus one minus one over two times 𝑥 plus one.

So far, our examples have had two unique factors in the denominator. Let’s see what would happen if instead we had three unique factors in the denominator.

Express 𝑥 squared minus two all over 𝑥 plus two times 𝑥 minus three multiplied by 𝑥 plus one in partial fractions.

We need to write this expression in partial fractions. And the first thing we always do is check the degree of the polynomial in our numerator and check the degree of the polynomial in our denominator. In our numerator, the highest power of 𝑥 is the term 𝑥 squared. So the degree of the polynomial in our numerator is two. In our denominator, when we multiply out our parentheses, we’ll see we’ll have 𝑥 cubed.

So the degree of the polynomial in our denominator is three. So our denominator is a higher-degree polynomial than our numerator. This means we can proceed with partial fractions. We see we have three unique roots in our denominator. And when our denominator has three unique roots, we need to split our fraction into three terms. By partial fractions, we get that our fraction is equal to 𝐴 over 𝑥 plus one plus 𝐵 over 𝑥 plus two plus 𝐶 over 𝑥 minus three for some constants 𝐴, 𝐵, and 𝐶.

We now want to add these three terms together. To do this, we need their denominators to match. Let’s start with our first term. Our first term has the denominator 𝑥 plus one. So we’ll multiply the numerator and the denominator of this fraction by 𝑥 plus two times 𝑥 minus three. So we multiply 𝐴 over 𝑥 plus one by 𝑥 plus two times 𝑥 minus three divided by 𝑥 plus two times 𝑥 minus three. We now want to rewrite the denominator in our second term in the same way. We’ll multiply the numerator and the denominator of this fraction by 𝑥 plus one times 𝑥 minus three. This time, for our second term, we get 𝐵 over 𝑥 plus two times 𝑥 plus one times 𝑥 minus three divided by 𝑥 plus one multiplied by 𝑥 minus three. We want to do the same with our third and final term. This time, we need to multiply the numerator and the denominator by 𝑥 plus one times 𝑥 plus two. And this gives us the following expression.

Now that we’ve rewritten our denominators to be equal, we can just add these terms together by adding the numerators together. When we do this, we get 𝐴 times 𝑥 plus two times 𝑥 minus three plus 𝐵 multiplied by 𝑥 plus one multiplied by 𝑥 minus three plus 𝐶 times 𝑥 plus one times 𝑥 plus two all divided by 𝑥 plus one times 𝑥 plus two times 𝑥 minus three. Now we can see the denominator on the left-hand side of our equation and the denominator on the right-hand side of our equation are equal. So for these two functions to be equal, the numerators must also be equal.

This means we can just equate our numerators. We get 𝑥 squared minus two is equal to 𝐴 times 𝑥 plus two times 𝑥 minus three plus 𝐵 multiplied by 𝑥 plus one times 𝑥 minus three plus 𝐶 times 𝑥 plus one multiplied by 𝑥 plus two. And remember, this has to be true for all values of 𝑥. So we can eliminate variables by substituting different values of 𝑥. For example, if we substitute 𝑥 is equal to negative two, then our coefficient of 𝐴 and our coefficient of 𝐶 are equal to zero.

Substituting 𝑥 is equal to negative two, on the left-hand side of our equation, we get negative two squared minus two. And remember, our coefficients of 𝐴 and 𝐶 will be zero, so we only need to substitute 𝑥 is equal to negative two to find our coefficient of 𝐵. So we’ll substitute 𝑥 is equal to negative two to find our coefficient of 𝐵. We get 𝐵 times negative two plus one multiplied by negative two minus three. Simplifying this equation, we get two is equal to five 𝐵. And then we just divide through by five to see that 𝐵 is equal to two-fifths.

We can do something similar to find the value of 𝐶. This time, we’ll substitute 𝑥 is equal to three. Now when we substitute this in, our coefficients of 𝐴 and 𝐵 will both be equal to zero. Substituting 𝑥 is equal to three and using the fact that our coefficients of 𝐴 and 𝐵 will be equal to zero, we get three squared minus two is equal to 𝐶 times three plus one times three plus two. Simplifying this expression, we get that seven is equal to 20𝐶, and then we can just solve this. We see that 𝐶 is equal to seven divided by 20.

Finally, we do exactly the same to find our value of 𝐴. This time we’re going to substitute 𝑥 is equal to negative one. Doing this and remembering our coefficients of 𝐵 and 𝐶 will now be equal to zero, we get negative one squared minus two is equal to 𝐴 times negative one plus two multiplied by negative one minus three. And then we simplify and solve this equation and we get that 𝐴 is equal to one-quarter.

The last thing we need to do to solve this equation is to substitute our values of 𝐴, 𝐵, and 𝐶 into our expression in partial fractions. So, by using 𝐴 is one-quarter, 𝐵 is two-fifths, and 𝐶 is seven over 20, we were able to express 𝑥 squared minus two divided by 𝑥 plus two times 𝑥 minus three times 𝑥 plus one in partial fractions. We got the expression one over four times 𝑥 plus one plus two over five times 𝑥 plus two plus seven over 20 times 𝑥 minus three.

The last thing we need to talk about is what happens if the polynomial in our numerator is of an equal or higher degree than the polynomial in our denominator.

Express 𝑥 squared minus five divided by 𝑥 squared minus four in a form which we can rewrite into partial fractions.

We’re given a quadratic divided by a quadratic. Since our numerator has an equal degree to our denominator, we can’t directly write this in partial fractions. So we’ll have to do something different. We’ll rewrite this by using algebraic division. We see that 𝑥 squared minus four goes into 𝑥 squared minus five one time. So it goes in once and then we need to subtract one times 𝑥 squared minus four. We have 𝑥 squared minus 𝑥 squared is equal to zero, and then negative five minus negative four is equal to negative one.

So 𝑥 squared minus four goes into 𝑥 squared minus five once with a remainder of one. But what does this actually mean? When we divide 𝑥 squared minus five by 𝑥 squared minus four, we get one and a remainder of one. So it’s equal to one plus one over 𝑥 squared minus four. And then, this is one plus a constant divided by a quadratic. So we can rewrite this by using partial fractions.

Let’s go over the key points of this video. The first thing we do when we’re using partial function decomposition is to check the degrees of the polynomial in our numerator and denominator. And if our numerator is of a higher or equal degree to our denominator, we can rewrite this fraction by using algebraic division. The next thing we need to do is factor the denominator fully. Then if all of our factors are unique and linear, we just write this as a sum of constants divided by each of our factors. Finally, we just add these fractions together and solve for our constants by substitution or by equating coefficients.