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

Lesson Explainer: Partial Fractions: Repeated Linear Factors Mathematics

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

The decomposition of an algebraic fraction into partial fractions is reversing the process of adding algebraic fractions. It is a way to simplify the denominators of algebraic fractions. We have seen how to apply this process when the denominator is the product of distinct linear factors; we split the algebraic fraction into the sum of algebraic fractions with each factor as a denominator. However, when dealing with repeated linear factors, we have to follow a different process.

To see why this is the case, letβs consider the decomposition of . We see that there is a repeated factor of in the denominator. If we tried to decompose this algebraic fraction into into the sum of algebraic fractions with each linear factor as a denominator, then we would have an expression of the form . However, adding these together does not give us a cubic polynomial in the denominator.

Instead, we might try the decomposition . If we add these expressions together, then we will obtain

For this to be equal to , the numerators must be equal. We have

This poses a problem. We note that for both sides of the equation to be equivalent, the corresponding coefficients must be equal. Equating the coefficients of yields ; however, equating the constant terms gives us . This contradiction means we cannot decompose the algebraic fraction into this form.

To remedy this problem, we can add an extra term:

Now, we can add the three algebraic fractions together by rewriting each to have the same denominator:

This expression must be equivalent to the fraction we are trying to decompose. Therefore, we have

Both sides have the same denominator, so their numerators must also be equal:

We can find the values of , , and by substitution and equating coefficients. We do this by choosing substitutions that set the factors to be zero.

Substituting into the equation yields

Substituting into the equation gives us

We note that we cannot find by setting factors to zero in this way since substituting or will result in the term with vanishing. However, since we have already found and , if we substitute in their values, then we have an equation with just that we can solve with substitution or equating coefficients.

Letβs find the value of by equating coefficients. We expand the brackets to obtain

Now, we can equate the coefficients of on both sides of the equation to find

Substituting into the equation gives

Hence, we can substitute these values of , , and into the equation to decompose the algebraic fraction into partial fractions:

In general, we have the following result.

Definition: Decomposition into Partial Fractions

If is an algebraic fraction where the degree of is lower than the degree of and is fully factored into linear factors, then we can decompose into partial fractions.

For each unique linear factor of , say, , we have a single algebraic fraction in the decomposition:

If has a repeated factor, say, is repeated twice, then we have 2 algebraic fractions in the decomposition: one for each power of the repeated root.

We can apply this to a polynomial of any degree; however, we will usually only be working with quadratics and cubics.

For example, if we have a repeated root in the quadratic in the denominator, say, , where the degree of is less than 2, then we can decompose this into partial fractions of the form for unknowns and .

If we have a repeated root and a nonrepeated root in a cubic in the denominator, say, , where and the degree of is less than 3, then we can decompose this into partial fractions of the form for unknowns , , and .

In our first example, we will consider a case where we have already been given the correct form of the partial fraction decomposition and just need to calculate the unknowns.

Example 1: Partial Fractions with Two Repeated Linear Factors

Find and such that .

Answer

We can find the values of and by first adding the algebraic fractions on the right-hand side of the equation together. To do this, we need them to have the same denominator. We can do this by multiplying by to get

Now that the algebraic fractions have the same denominator, we can add them together by adding their numerators:

This must be equal to the left-hand side of our original equation. So, we have

The denominators of both sides of the equation are equal, so the numerators must also be equal. This gives us

We can find the values of and by substitution.

Substituting into the equation yields

We can then find the value of by substitution or by equating coefficients. Letβs expand the brackets to obtain

Now, we can equate the coefficients of on both sides of the equation to find

Hence,

Now that we have worked through an example where we are given the correct decomposition, letβs try one where we need to work it out ourselves. In particular, we will consider a fraction with a cubic denominator that has been factored to give us one repeated root.

Example 2: Partial Fractions with Repeated Linear Factors

Determine the partial fraction decomposition of .

Answer

We first note that we have a factored cubic in the denominator and that it has a repeated root. The degree of the polynomial in the numerator is less than that in the denominator.

Recall that to decompose something into partial fractions, we will need to rewrite it as a sum of fractions with linear factors in the denominators.

Since one of the roots is repeated, we recall that the partial fraction decomposition will be different from the usual case. As there is an in the denominator, we will need one fraction of the form and one of the form . The third one will contain the other factor, that is, . Thus, we have for some unknown values of , , and .

We can determine the values of , , and by adding the algebraic fractions together on the right-hand side of the equation.

To add the algebraic fractions together, we need to rewrite them to have the same denominator:

Now that the algebraic fractions have the same denominator, we can add them together by adding their numerators:

This expression must be equal to the algebraic fraction given to us in the question:

The denominators of both sides of the equation are equal, so the numerators must also be equal. This gives us

We can determine the values of , , and by substitution and equating coefficients.

Substituting into the equation yields

Substituting into the equation and solving it gives us

To find the value of , we want to equate coefficients. Letβs start by expanding the brackets to obtain

We see that there is no term on the left-hand side of the equation, so the coefficient of is 0. On the right-hand side of the equation, there are two terms that we can combine to make . Equating the coefficients of gives us

Substituting into this equation and solving it gives

Hence,

It is worth noting that there are shortcuts to make the process of partial fraction decomposition easier. For instance, in the previous example, we saw that we wanted to decompose the algebraic fraction into the form

Instead of cross multiplying and adding the fractions on the right-hand side of the equation together, we can multiply each of the numerators by the denominator , where we remove the factors from the denominator. This is equivalent to just multiplying the equation through by :

In our next example, we will consider a partial fraction decomposition where the expression has a linear numerator.

Example 3: Partial Fractions with Repeated Linear Factors

Express in partial fractions.

Answer

We first note that we have a factored cubic in the denominator with a repeated root and that the degree of the polynomial in the numerator is less than that in the denominator. We recall that this means that we can use partial fractions to rewrite the expression in terms of fractions with the linear factors and all of the powers of the repeated factors in the denominator: for some unknown values of , , and .

We can determine the values of these unknowns by adding the algebraic fractions together on the right-hand side of the equation.

To do this, we need to rewrite the algebraic fractions to have the same denominator:

Now that the algebraic fractions have the same denominator, we can add them together by adding their numerators:

This expression must be equal to the algebraic fraction we are trying to decompose:

We can now note that the denominators of both sides of the equation are equal, so the numerators must also be equal:

We can determine the values of , , and by substitution and equating coefficients.

Substituting into the equation and solving it yields

Substituting into the equation and solving it gives us

To find the value of , we want to equate coefficients. Letβs start by expanding the brackets to obtain

We see that there is no term on the left-hand side of the equation, so the coefficient of is 0. On the right-hand side of the equation, the coefficient of is . So, we have .

Hence,

In our next example, we will determine the partial fraction decomposition of an algebraic fraction with an unfactored quadratic denominator. This means that we will need to perform the extra step of factoring the denominator before we can proceed with the usual method of decomposition.

Example 4: Factoring Denominators and Applying Partial Fractions with Two Repeated Linear Factors

Express in partial fractions.

Answer

We first recall that to express an algebraic fraction in partial fractions, we first need to fully factorize the polynomial in the denominator. To factorize the quadratic in the denominator, we need to find two numbers whose product is 9 and whose sum is . By considering the factors of 9, we see that these are and . Thus,

We might have also noticed that the above quadratic is a perfect square, meaning that we could have factored it using the formula.

We now note that the denominator has a repeated root. We recall that this means that we can use partial fractions to rewrite the expression in terms of fractions with the powers of the repeated linear factors as denominators: for some unknown values and .

We can determine the values of these unknowns by adding the algebraic fractions together on the right-hand side of the equation.

To do this, we need to rewrite the algebraic fractions to have the same denominator:

This expression must be equal to the algebraic fraction we are trying to decompose. Therefore, we have

We can now note that the denominators of both sides of the equation are equal, so the numerators must also be equal:

We can determine the values of and by substitution and equating coefficients.

Substituting into the equation and solving it yields

To find the value of , we want to equate coefficients. We can expand the brackets to obtain

The coefficient of on the left-hand side of the equation is , and on the right-hand side, the coefficient of is . Equating these coefficients gives us

Hence,

In our final example, we will once again consider the partial fraction decomposition of an algebraic fraction with an unfactored denominator. This time, however, we will need to factor a cubic expression.

Example 5: Factoring Denominators and Applying Partial Fractions with Repeated Linear Factors

Decompose into partial fractions.

Answer

We first recall that to express an algebraic fraction in partial fractions, we first need to fully factorize the polynomial in the denominator. We see that both terms in the denominator share a factor of , so we have

Thus,

We now note that we have a factored cubic in the denominator with a repeated root and that the degree of the polynomial in the numerator is less than that in the denominator. We recall that this means that we can use partial fractions to rewrite the expression in terms of fractions with the linear factors and all of the powers of the repeated factors as denominators: for some unknown constants , , and .

We can determine the values of these unknowns by adding the algebraic fractions together on the right-hand side of the equation. However, we will instead multiply through by . We obtain

We can determine the values of , , and by substitution and equating coefficients.

Substituting into the equation and solving it yields

Substituting into the equation and solving it gives us

To find the value of , we want to equate coefficients. Letβs start by expanding the brackets to obtain

The coefficient of on the left-hand side of the equation is 2. On the right-hand side of the equation, there are two terms that we can combine to make . Equating the coefficients of gives us

Substituting into this equation and solving it gives

Hence,

Letβs finish by recapping some of the important points from this explainer.

Key Points

• When decomposing an algebraic fraction with a linear root that is repeated twice, we will obtain two algebraic fractions in the decomposition whose denominators are each of the powers of the linear factor.
• We can only directly apply partial fraction decomposition when the numerator has a lower degree than the denominator.
• There are two methods we can use to find the unknown constants after the decomposition of an algebraic fraction into partial fractions: substitution and equating coefficients. It is important to be comfortable with us using both methods.
• When dealing with partial fraction decomposition with repeated linear roots, we can find the values of the constants using a combination of substitution and equating coefficients.
• Sometimes, we may need to factor the denominator in the given algebraic fraction so that we can decompose it into partial fractions.

Download the Nagwa Classes App

Attend sessions, chat with your teacher and class, and access class-specific questions. Download the Nagwa Classes app today!

Download for Desktop

Windows macOS Intel macOS Apple Silicon

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy