Lesson Video: Partial Fractions: Repeated Linear Factors | Nagwa Lesson Video: Partial Fractions: Repeated Linear Factors | Nagwa

# Lesson Video: Partial Fractions: Repeated Linear Factors Mathematics

In this video, we will learn how to decompose rational expressions into partial fractions when the denominator has repeated linear factors.

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### Video Transcript

In this video, weβll learn how to decompose rational expressions into partial fractions when the denominator has repeated linear factors. We begin by recalling the fact that a fraction or a rational function with two or more distinct linear factors in its denominator and whose denominator has a higher order than the polynomial in its numerator can be split into two or more separate fractions with linear denominators. We essentially reverse the process of adding algebraic fractions to achieve this, either by using substitution or equating coefficients.

We use partial fraction decomposition in integration and in evaluating binomial expansions. So itβs a really important skill to master. Further, the method of partial fractions can be used when there are repeated linear factors in the denominator, such as in the case of two π₯ plus one over π₯ minus three times π₯ plus one cubed, though it does look slightly different. When dealing with a repeated linear factor, here thatβs π₯ plus one since itβs being cubed, we follow the usual steps for dealing with nonrepeated linear factors. This time, though, if π is the exponent of the repeated linear factor on the denominator, we write π rational expressions, each of which has that linear factor raised to increasing exponents in the denominator.

Note of course that for linear factors, all our numerators will be constant. So for our nonrepeated factor, π₯ minus three, we write π΄ over π₯ minus three. Then, for our repeated factor since the exponent is three, we know we need three rational expressions. Their denominators will be π₯ plus one in increasing exponent. So we have π΅ over π₯ plus one, πΆ over π₯ plus one squared, and π· over π₯ plus one cubed. Letβs begin by looking at an example that demonstrates the key process.

Find π΄ and π΅ such that two π₯ over π₯ minus three squared is equal to π΄ over π₯ minus three plus π΅ over π₯ minus three squared.

This question is actually just asking us to write the expression two π₯ over π₯ minus three squared in partial fraction form. It has a repeated linear factor in its denominator. The repeated linear factor is π₯ minus three. And we know itβs repeated because itβs being squared. The process for performing partial fraction decomposition in this case is as follows. We begin by checking that the order or the degree of the polynomial in our denominator is higher than the order of the polynomial in our numerator. If itβs not, weβll need to use polynomial long division before splitting our remainder into partial fraction form.

Well, if we were to distribute the parentheses on our denominator, weβd see we have a quadratic expression. Its highest exponent is two. So the order of the polynomial on our denominator is two. We have a linear expression on the numerator. Its order is one. The order or degree of our polynomial on the denominator is indeed higher than that of our numerator. So we move on to step two. Step two is to factor the denominator if necessary. π₯ minus three squared is already in factored form, so we move on. Then we deal with any nonrepeated linear factors by following the usual process. There are in fact no nonrepeated factors in our denominator, so weβre going to move on.

This is where the process differs from dealing with nonrepeated factors. We say that if π is the exponent of the repeated factor on the denominator, weβre going to write π rational expressions. Each of which has that factor raised to increasing exponents in its denominator. Note that we donβt actually necessarily need to use increasing exponents. We could use decreasing exponents. But by always using increasing exponents in our denominators, we ensure that we donβt lose any of our fractions. Our repeated factor, remember, was π₯ minus three, and itβs raised to the power of two. So π is equal to two. And this means we need two rational expressions when splitting it into partial fractions.

The denominator will be π₯ minus three to increasing exponents. Well, this has actually been done for us. We have π₯ minus three, which is essentially π₯ minus three to the power of one. And we have π₯ minus three squared. Note that the numerator of these fractions will be a real constant. And thatβs because the denominator is a linear factor. So thatβs step four completed. Whatβs the final step? We do something called multiplying up. And what we mean by this is we add these two fractions by creating a common denominator. And that denominator will match the denominator of the original fraction. Once weβve done that, weβre able to evaluate the constants, either by using substitution or equating coefficients.

So letβs see what step five looks like. We have two π₯ over π₯ minus three squared is equal to some constant π΄ over π₯ minus three plus some constant π΅ over π₯ minus three squared. Note that this is sometimes written as an identity symbol. And this is because this expression is true for all values of π₯. In this question, though, weβre going to keep it as an equal sign and just bear this in mind. Remember, we want to add the fractions π΄ over π₯ minus three plus π΅ over π₯ minus three squared. And we do so by creating a common denominator. But this common denominator will always match the denominator on the left-hand side. Note that the denominator of this fraction, π΅ over π₯ minus three squared, is already of this form. And so weβre actually only going to look at this first fraction, π΄ over π₯ minus three.

To achieve a denominator of π₯ minus three squared, we need to multiply the denominator of this fraction by π₯ minus three. But of course, whatever we do to the bottom, we have to do to the top. So weβre going to multiply both the numerator and denominator by π₯ minus three. And so our first fraction becomes π΄ times π₯ minus three over π₯ minus three squared. We now notice that the denominators of our two fractions are equal. And so weβre able to add their numerators. That gives us π΄ times π₯ minus three plus π΅ all over π₯ minus three squared. This, of course, is still equal to the fraction two π₯ over π₯ minus three squared. And now, we note that the denominators of our fractions are equal.

So for the two rational functions to be equal, their numerators must also be equal. In other words, two π₯ must be equal to π΄ times π₯ minus three plus π΅ for all values of π₯. We have two ways to evaluate the constants. The first way is to use substitution. Letβs see what that looks like. We choose values of π₯ that match the roots of our denominator. So the root of the equation π₯ minus three squared equals zero is π₯ equals three. We substitute these in in turn. So we let π₯ be equal to three, and the left-hand side of our equation becomes two times three. The right-hand side becomes π΄ times three minus three plus π΅.

But of course, three minus three is zero. And note that by setting π₯ equal to three, weβve obtained an equation where the coefficient of π΄ is zero. And so we have an equation purely in π΅. That is, π΅ is equal to six. Now we know that our identity is true for all values of π₯. So letβs replace π΅ with six in our original equation. We get two π₯ equals π΄ times π₯ minus three plus six. Weβre now going to choose any value of π₯ and substitute that in. Letβs choose a simple one like π₯ equals zero. When we do, we get two times zero equals π΄ times zero minus three plus six. Simplifying, our equation becomes zero equals negative three π΄ plus six. And we solve for π΄ by adding three π΄ to both sides and then dividing through by three. We find π΄ is equal to two.

So we found π΅ is equal to six and π΄ is equal to two. Letβs very briefly look at the other way we couldβve found these values. We go back to our earlier equation. And we distribute the parentheses on the right-hand side. That gives us π΄π₯ minus three π΄ plus π΅. In this method, we equate coefficients. We compare the coefficients of our various exponents of π₯. Letβs begin by comparing the coefficients of π₯ to the power of one on both sides or simply π₯. On the left-hand side, we see the coefficient of π₯ is two. On the right-hand side, we see the coefficient of π₯ is π΄. And so we formed an equation. Itβs π΄ is equal to two.

Weβll now compare the coefficients of π₯ to the power of zero, in other words, the constant values. On the left-hand side, there are no constants. So the coefficient of π₯ to the power of zero is zero. On the right-hand side, the coefficients of π₯ to the power of zero, the constants, are negative three π΄ plus π΅. But, of course, we just evaluated π΄ to be two. So we get zero equals negative three times two plus π΅. That is, zero equals negative six plus π΅. And we solve for π΅ by adding six to both sides. Once again, we find π΄ is equal to two and π΅ is equal to six.

Note that both of these methods are equally valid. In an example like this, equating coefficients can be nice and efficient. However, with more complicated examples, we might end up with a pair of simultaneous equations or even three simultaneous equations in three unknowns. In that case, we would use substitution or a mixture of both.

Letβs now look at an example which involves more than two factors.

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

Here, weβre looking to determine the partial fraction decomposition of a fraction which has a repeated linear factor in its denominator. That repeated factor is π₯ minus one. And we know itβs repeated because itβs being squared. The process for performing partial fraction decomposition in this case is as follows. The first step is to check that the order of the polynomial in the denominator is greater than that of the numerator. While the order or degree of the polynomial in our denominator is full, since distributing would yield a highest power of π₯ of four. The order of our numerator is two. And so we can say that yes, the order of the denominator or the degree of the denominator is greater than that of the numerator.

Remember, if this wasnβt the case, weβd need to use polynomial long division before splitting the remainder into partial fraction form. Next, we factor our denominator if necessary. Well, this is already in factored form, and so we move on. We then look at our nonrepeated factors. And we follow the usual process. We essentially perform the reverse to adding algebraic fractions. Our nonrepeated linear factors are π₯ and π₯ minus three. So we begin the process by splitting our fraction into π΄ over π₯ and π΅ over π₯ minus three. Remember, π΄ and π΅ are simply constants. And this is because we have linear factors on our denominator.

We then move on to our repeated factor. And we say that imagine π is the exponent of the repeated factor. Weβre going to write π rational expressions, each of which has that factor raised to increasing exponents as its denominator. Well, our repeated factor is π₯ plus one, and itβs raised to the power of two. In other words, π is equal to two. So we want π rational expressions with π₯ plus one raised to increasing exponents. Thatβs πΆ over π₯ plus one plus π· over π₯ plus one squared. Note once again that since these are linear denominators, their numerators are simply constants. Now we have this in the correct form, letβs clear some space and perform the final steps.

The final step is to do something called multiplying up. Essentially, we take our fractions on the right-hand side and we add them by creating a common denominator. And that matches the denominator on the left-hand side. And then we evaluate the constants. So letβs perform that first step. We know the denominator is going to be π₯ times π₯ minus three times π₯ plus one squared. So what are we going to multiply each of our fractions by? For the first fraction, weβre going to need to multiply both the numerator and the denominator by π₯ minus three times π₯ plus one squared. That gives us π΄ times π₯ minus three times π₯ plus one squared over the denominator we required.

Weβll now look at our second fraction. We see that weβll need to multiply this by π₯ times π₯ plus one squared to achieve that denominator. And so we get a numerator of π΅π₯ times π₯ plus one squared. Weβll repeat this process for the last two fractions. Weβre going to need to multiply this third one by π₯ times π₯ minus three times π₯ plus one and this fourth one by π₯ times π₯ minus three. And so the numerators of our final two fractions are πΆπ₯ times π₯ minus three times π₯ plus one and π·π₯ times π₯ minus three. Note now that the denominators of each of our fractions are equal. And so we can add their numerators. And in doing so, we get the expression shown.

Note of course that this is still equal to our original fraction. And we might sometimes use an identity symbol to show that this is true for all values of π₯. Weβre going to leave this as an equal sign, but bear this in mind. Now, the whole purpose of doing what we just did is because weβve created two fractions whose denominators are equal. And the fractions themselves are equal for all values of π₯. So this must mean that their numerators must also be equal. That is, π₯ squared plus π₯ plus one must be equal to π΄ times π₯ minus three times π₯ plus one squared plus π΅π₯ times π₯ plus one squared plus πΆπ₯ times π₯ minus three times π₯ plus one plus π·π₯ times π₯ minus three.

Now to evaluate the constants, we have two options. We could distribute all of our parentheses and equate coefficients. The problem is because weβre looking to find the value of four constants, weβre going to end up having a rather nasty set of simultaneous equations. And so instead, weβre going to perform substitution. We look for the roots of the denominator of our original fraction, and we substitute these in. One of the roots to the equation π₯ times π₯ minus three times π₯ plus one squared equals zero is π₯ equals zero. So letβs substitute zero into this equation. On the left-hand side, we get zero squared plus zero plus one, which is simply equal to one. On the right-hand side, we get π΄ times negative three times one squared. But then, π΅ times π₯ times π₯ plus one squared becomes π΅ times zero, which is zero.

Similarly, this term becomes zero and this term becomes zero. So the whole purpose of substituting π₯ equals zero into this equation was to eliminate the terms involving π΅, πΆ, and π· and form an equation purely in π΄. That equation simplifies to one equals negative three π΄. And we solve for π΄ by dividing through by three to get π΄ equals negative one-third. Letβs repeat this process with a different root. Another root is π₯ equals three, so letβs substitute that in. On the left-hand side, letting π₯ equal to three gives us 13. Then on the right, this term, this term, and this term all contain π₯ minus three. So three minus three is zero, and they essentially disappear. And our equation becomes 13 equals π΅ times three times four squared or 13 equals 48π΅. Solving for π΅ gives us 13 over 48.

Thereβs one more root we can try, that is, π₯ equals negative one. This time, letting π₯ equal negative one eliminates this term, this term, and this term. And their equation becomes one equals π· times negative one times negative four or one equals four π·. And so we find π· is equal to a quarter. But how do we work out the value of πΆ? Well, this time we can use something called equating coefficients. Letβs imagine we were going to distribute the parentheses on the left- and right-hand side. What would the coefficients of π₯ cubed be? Well, on the right-hand side, there are no π₯ cubed terms, so itβs zero. If we were to distribute the parentheses on the right-hand side, weβd get π΄ lots of π₯ cubed plus π΅ lots of π₯ cubed plus πΆ lots of π₯ cubed. So we can say that zero must be equal to π΄ plus π΅ plus πΆ.

But of course, we evaluated π΄ to be negative one-third and π΅ to be equal to 13 over 48. Solving this equation for πΆ, and we find itβs equal to three over 48 or one sixteenth. All thatβs left is to replace π΄, π΅, πΆ, and π· with their values. Weβll begin with the largest exponent in the denominator. Thatβs this last term with π· on it. Itβs a quarter over π₯ plus one squared, which is one over four times π₯ plus one squared. Next, dealing with πΆ, and we get one sixteenth over π₯ plus one, which is one over 16 times π₯ plus one. Then π΄ over π₯ becomes negative one-third over π₯, which is negative one over three π₯. And finally, we have 13 over 48 over π₯ minus three, which we can write as shown. And weβve determined the partial fraction decomposition of our rational function.

In this video, weβve learned that the process for performing partial fraction decomposition with our repeated linear factors in the denominator is roughly the same as when there are no repeated linear factors. However, once weβve dealt with any nonrepeated factors, we say that if π is the exponent of the repeated factor, we write π rational expressions, each of which has that factor raised to increasing exponents as their denominators. We then multiply up and use substitution or equating coefficients to calculate the value of the constants.

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