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, let’s consider the decomposition of 3𝑥1𝑥(𝑥1). We see that there is a repeated factor of 𝑥 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 𝐴𝑥+𝐵𝑥1. However, adding these together does not give us a cubic polynomial in the denominator.

Instead, we might try the decomposition 𝐴𝑥+𝐵𝑥1. If we add these expressions together, then we will obtain 𝐴𝑥+𝐵𝑥1=𝐴(𝑥1)𝑥(𝑥1)+𝐵𝑥𝑥(𝑥1)=𝐴(𝑥1)+𝐵𝑥𝑥(𝑥1).

For this to be equal to 3𝑥1𝑥(𝑥1), the numerators must be equal. We have 3𝑥1=𝐴(𝑥1)+𝐵𝑥=𝐵𝑥+𝐴𝑥𝐴.

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 𝑥 yields 𝐴=0; however, equating the constant terms gives us 𝐴=1. This contradiction means we cannot decompose the algebraic fraction into this form.

To remedy this problem, we can add an extra term: 3𝑥1𝑥(𝑥1)=𝐴𝑥+𝐵𝑥+𝐶𝑥1.

Now, we can add the three algebraic fractions together by rewriting each to have the same denominator: 𝐴𝑥+𝐵𝑥+𝐶𝑥1=𝐴𝑥×𝑥(𝑥1)𝑥(𝑥1)+𝐵𝑥×𝑥1𝑥1+𝐶𝑥1×𝑥𝑥=𝐴𝑥(𝑥1)𝑥(𝑥1)+𝐵(𝑥1)𝑥(𝑥1)+𝐶𝑥𝑥(𝑥1)=𝐴𝑥(𝑥1)+𝐵(𝑥1)+𝐶𝑥𝑥(𝑥1).

This expression must be equivalent to the fraction we are trying to decompose. Therefore, we have 3𝑥1𝑥(𝑥1)=𝐴𝑥(𝑥1)+𝐵(𝑥1)+𝐶𝑥𝑥(𝑥1).

Both sides have the same denominator, so their numerators must also be equal: 3𝑥1=𝐴𝑥(𝑥1)+𝐵(𝑥1)+𝐶𝑥.

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

Substituting 𝑥=1 into the equation yields 3(1)1=𝐴(1)(11)+𝐵(11)+𝐶(1)2=𝐶.

Substituting 𝑥=0 into the equation gives us 3(0)1=𝐴(0)(01)+𝐵(01)+𝐶(0)1=𝐵𝐵=1.

We note that we cannot find 𝐴 by setting factors to zero in this way since substituting 𝑥=0 or 𝑥=1 will result in the term with 𝐴 vanishing. However, since we have already found 𝐵 and 𝐶, if we substitute in their values, then we have an equation with just 𝐴 that we can solve with substitution or equating coefficients.

Let’s find the value of 𝐴 by equating coefficients. We expand the brackets to obtain 3𝑥1=𝐴𝑥𝐴+𝐵𝑥𝐵+𝐶𝑥.

Now, we can equate the coefficients of 𝑥 on both sides of the equation to find 3=𝐴+𝐶.

Substituting 𝐶=2 into the equation gives 3=𝐴+2𝐴=1.

Hence, we can substitute these values of 𝐴, 𝐵, and 𝐶 into the equation to decompose the algebraic fraction into partial fractions: 3𝑥1𝑥(𝑥1)=1𝑥+1𝑥+2𝑥1.

In general, we have the following result.

Definition: Decomposition into Partial Fractions

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

For each unique linear factor of 𝑄(𝑥), say, 𝑥𝑎, we have a single algebraic fraction in the decomposition: 𝐴𝑥𝑎.

If 𝑄(𝑥) has a repeated factor, say, 𝑥𝑏 is repeated twice, then we have 2 algebraic fractions in the decomposition: 𝐵𝑥𝑏,𝐵(𝑥𝑏), 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, 𝑃(𝑥)(𝑥𝑎), where the degree of 𝑃(𝑥) is less than 2, then we can decompose this into partial fractions of the form 𝑃(𝑥)(𝑥𝑎)=𝐴𝑥𝑎+𝐵(𝑥𝑎), for unknowns 𝐴 and 𝐵.

If we have a repeated root and a nonrepeated root in a cubic in the denominator, say, 𝑃(𝑥)(𝑥𝑎)(𝑥𝑏), where 𝑎𝑏 and the degree of 𝑃(𝑥) is less than 3, then we can decompose this into partial fractions of the form 𝑃(𝑥)(𝑥𝑎)(𝑥𝑏)=𝐴𝑥𝑎+𝐵(𝑥𝑎)+𝐶𝑥𝑏, for unknowns 𝐴, 𝐵, and 𝐶.

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 𝐴 and 𝐵 such that 2𝑥(𝑥3)=𝐴𝑥3+𝐵(𝑥3).

Answer

We can find the values of 𝐴 and 𝐵 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 𝐴𝑥3by 𝑥3𝑥3 to get 𝐴𝑥3+𝐵(𝑥3)=𝐴𝑥3×𝑥3𝑥3+𝐵(𝑥3)=𝐴(𝑥3)(𝑥3)+𝐵(𝑥3).

Now that the algebraic fractions have the same denominator, we can add them together by adding their numerators: 𝐴(𝑥3)(𝑥3)+𝐵(𝑥3)=𝐴(𝑥3)+𝐵(𝑥3).

This must be equal to the left-hand side of our original equation. So, we have 2𝑥(𝑥3)=𝐴(𝑥3)+𝐵(𝑥3).

The denominators of both sides of the equation are equal, so the numerators must also be equal. This gives us 2𝑥=𝐴(𝑥3)+𝐵.

We can find the values of 𝐴 and 𝐵 by substitution.

Substituting 𝑥=3 into the equation yields 2(3)=𝐴(33)+𝐵6=𝐵.

We can then find the value of 𝐴 by substitution or by equating coefficients. Let’s expand the brackets to obtain 2𝑥=𝐴𝑥3𝐴+𝐵.

Now, we can equate the coefficients of 𝑥 on both sides of the equation to find 𝐴=2.

Hence, 2𝑥(𝑥3)=2𝑥3+6(𝑥3).

Now that we have worked through an example where we are given the correct decomposition, let’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 1𝑥(𝑥1).

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 𝑥 in the denominator, we will need one fraction of the form 𝐴𝑥 and one of the form 𝐵𝑥. The third one will contain the other factor, that is, 𝐶𝑥1. Thus, we have 1𝑥(𝑥1)=𝐴𝑥+𝐵𝑥+𝐶𝑥1, for some unknown values of 𝐴, 𝐵, and 𝐶.

We can determine the values of 𝐴, 𝐵, and 𝐶 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: 𝐴𝑥+𝐵𝑥+𝐶𝑥1=𝐴𝑥×𝑥(𝑥1)𝑥(𝑥1)+𝐵𝑥×(𝑥1)(𝑥1)+𝐶𝑥1×𝑥𝑥=𝐴𝑥(𝑥1)𝑥(𝑥1)+𝐵(𝑥1)𝑥(𝑥1)+𝐶𝑥𝑥(𝑥1).

Now that the algebraic fractions have the same denominator, we can add them together by adding their numerators: 𝐴𝑥(𝑥1)𝑥(𝑥1)+𝐵(𝑥1)𝑥(𝑥1)+𝐶𝑥𝑥(𝑥1)=𝐴𝑥(𝑥1)+𝐵(𝑥1)+𝐶𝑥𝑥(𝑥1).

This expression must be equal to the algebraic fraction given to us in the question: 1𝑥(𝑥1)=𝐴𝑥(𝑥1)+𝐵(𝑥1)+𝐶𝑥𝑥(𝑥1).

The denominators of both sides of the equation are equal, so the numerators must also be equal. This gives us 1=𝐴𝑥(𝑥1)+𝐵(𝑥1)+𝐶𝑥.

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

Substituting 𝑥=1 into the equation yields 1=𝐴(1)(11)+𝐵(11)+𝐶(1)1=𝐶.

Substituting 𝑥=0 into the equation and solving it gives us 1=𝐴(0)(01)+𝐵(01)+𝐶(0)1=𝐵𝐵=1.

To find the value of 𝐴, we want to equate coefficients. Let’s start by expanding the brackets to obtain 1=𝐴𝑥(𝑥1)+𝐵(𝑥1)+𝐶𝑥1=𝐴𝑥𝐴𝑥+𝐵𝑥𝐵+𝐶𝑥.

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

Substituting 𝐶=1 into this equation and solving it gives 0=𝐴+(1)𝐴=1.

Hence, 1𝑥(𝑥1)=1𝑥+1𝑥1𝑥1.

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 1𝑥(𝑥1)=𝐴𝑥+𝐵𝑥+𝐶𝑥1.

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 𝑥(𝑥1), where we remove the factors from the denominator. This is equivalent to just multiplying the equation through by 𝑥(𝑥1): 1=𝐴𝑥(𝑥1)𝑥+𝐵𝑥(𝑥1)𝑥+𝐶𝑥(𝑥1)𝑥11=𝐴𝑥(𝑥1)+𝐵(𝑥1)+𝐶𝑥.

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 5𝑥+1(𝑥+1)(𝑥2) 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: 5𝑥+1(𝑥+1)(𝑥2)=𝐴𝑥+1+𝐵(𝑥+1)+𝐶𝑥2, for some unknown values of 𝐴, 𝐵, and 𝐶.

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: 𝐴𝑥+1+𝐵(𝑥+1)+𝐶𝑥2=𝐴𝑥+1×(𝑥+1)(𝑥2)(𝑥+1)(𝑥2)+𝐵(𝑥+1)×𝑥2𝑥2+𝐶𝑥2×(𝑥+1)(𝑥+1)=𝐴(𝑥+1)(𝑥2)(𝑥+1)(𝑥2)+𝐵(𝑥2)(𝑥+1)(𝑥2)+𝐶(𝑥+1)(𝑥+1)(𝑥2).

Now that the algebraic fractions have the same denominator, we can add them together by adding their numerators: 𝐴(𝑥+1)(𝑥2)(𝑥+1)(𝑥2)+𝐵(𝑥2)(𝑥+1)(𝑥2)+𝐶(𝑥+1)(𝑥+1)(𝑥2)=𝐴(𝑥+1)(𝑥2)+𝐵(𝑥2)+𝐶(𝑥+1)(𝑥+1)(𝑥2).

This expression must be equal to the algebraic fraction we are trying to decompose: 5𝑥+1(𝑥+1)(𝑥2)=𝐴(𝑥+1)(𝑥2)+𝐵(𝑥2)+𝐶(𝑥+1)(𝑥+1)(𝑥2).

We can now note that the denominators of both sides of the equation are equal, so the numerators must also be equal: 5𝑥+1=𝐴(𝑥+1)(𝑥2)+𝐵(𝑥2)+𝐶(𝑥+1).

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

Substituting 𝑥=1 into the equation and solving it yields 5(1)+1=𝐴(1+1)(12)+𝐵(12)+𝐶(1+1)5+1=3𝐵𝐵=2.

Substituting 𝑥=2 into the equation and solving it gives us 5(2)+1=𝐴(2+1)(22)+𝐵(22)+𝐶(2+1)10+1=9𝐶𝐶=1.

To find the value of 𝐴, we want to equate coefficients. Let’s start by expanding the brackets to obtain 5𝑥+1=𝐴(𝑥+1)(𝑥2)2(𝑥2)(𝑥+1)=𝐴𝑥𝑥22(𝑥2)𝑥+2𝑥+1=𝐴𝑥𝐴𝑥2𝐴2𝑥+4𝑥2𝑥1=(𝐴1)𝑥+(𝐴4)𝑥2𝐴+3.

We see that there is no 𝑥 term on the left-hand side of the equation, so the coefficient of 𝑥 is 0. On the right-hand side of the equation, the coefficient of 𝑥 is 𝐴1. So, we have 𝐴=1.

Hence, 5𝑥+1(𝑥+1)(𝑥2)=1𝑥+12(𝑥+1)1𝑥2.

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 𝑥+5𝑥6𝑥+9 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 6. By considering the factors of 9, we see that these are 3 and 3. Thus, 𝑥6𝑥+9=(𝑥3).

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: 𝑥+5𝑥6𝑥+9=𝐴𝑥3+𝐵(𝑥3), for some unknown values 𝐴 and 𝐵.

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: 𝐴𝑥3+𝐵(𝑥3)=𝐴𝑥3×𝑥3𝑥3+𝐵(𝑥3)=𝐴(𝑥3)(𝑥3)+𝐵(𝑥3)=𝐴(𝑥3)+𝐵(𝑥3).

This expression must be equal to the algebraic fraction we are trying to decompose. Therefore, we have 𝑥+5𝑥6𝑥+9=𝐴(𝑥3)+𝐵(𝑥3).

We can now note that the denominators of both sides of the equation are equal, so the numerators must also be equal: 𝑥+5=𝐴(𝑥3)+𝐵.

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

Substituting 𝑥=3 into the equation and solving it yields 3+5=𝐴(33)+𝐵2=0𝐴+𝐵𝐵=2.

To find the value of 𝐴, we want to equate coefficients. We can expand the brackets to obtain 𝑥+5=𝐴(𝑥3)+𝐵𝑥+5=𝐴𝑥3𝐴+𝐵.

The coefficient of 𝑥 on the left-hand side of the equation is 1, and on the right-hand side, the coefficient of 𝑥 is 𝐴. Equating these coefficients gives us 𝐴=1.

Hence, 𝑥+5𝑥6𝑥+9=1𝑥3+2(𝑥3).

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 2𝑥+4𝑥+2𝑥 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 𝑥, so we have 𝑥+2𝑥=𝑥(𝑥+2).

Thus, 2𝑥+4𝑥+2𝑥=2𝑥+4𝑥(𝑥+2).

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: 2𝑥+4𝑥(𝑥+2)=𝐴𝑥+𝐵𝑥+𝐶𝑥+2, for some unknown constants 𝐴, 𝐵, and 𝐶.

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 𝑥(𝑥+2). We obtain 2𝑥+4=𝐴𝑥(𝑥+2)𝑥+𝐵𝑥(𝑥+2)𝑥+𝐶𝑥(𝑥+2)𝑥+22𝑥+4=𝐴𝑥(𝑥+2)+𝐵(𝑥+2)+𝐶𝑥.

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

Substituting 𝑥=0 into the equation and solving it yields 2(0)+4=𝐴(0)(0+2)+𝐵(0+2)+𝐶(0)4=2𝐵𝐵=2.

Substituting 𝑥=2 into the equation and solving it gives us 2(2)2+4=𝐴(2)(2+2)+𝐵(2+2)+𝐶(2)12=4𝐶𝐶=3.

To find the value of 𝐴, we want to equate coefficients. Let’s start by expanding the brackets to obtain 2𝑥+4=𝐴𝑥(𝑥+2)+𝐵(𝑥+2)+𝐶𝑥2𝑥+4=𝐴𝑥+2𝐴𝑥+𝐵𝑥+2𝐵+𝐶𝑥.

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

Substituting 𝐶=3 into this equation and solving it gives 2=𝐴+3𝐴=1.

Hence, 2𝑥+4𝑥+2𝑥=1𝑥+2𝑥+3𝑥+2.

Let’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.

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