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

Lesson Explainer: Partial Fractions: Nonrepeated Linear Factors Mathematics

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

In order to understand how we can decompose an algebraic fraction into partial fractions, we first need to consider how we add and subtract algebraic fractions. Consider the following sum of two algebraic fractions: 1𝑥+1+2𝑥3.

We recall that we add these together by rewriting each fraction to have the same denominator. We can do this by cross multiplication: 1𝑥+1+2𝑥3=1×(𝑥3)(𝑥+1)(𝑥3)+2×(𝑥+1)(𝑥3)(𝑥+1).

Now that the algebraic fractions have the same denominator, we can add them together by combining the numerators. This gives us 1×(𝑥3)(𝑥+1)(𝑥3)+2×(𝑥+1)(𝑥3)(𝑥+1)=(𝑥3)+2(𝑥+1)(𝑥+1)(𝑥3)=𝑥3+2𝑥+2(𝑥+1)(𝑥3)=3𝑥1(𝑥+1)(𝑥3).

Hence, we have shown that 1𝑥+1+2𝑥3=3𝑥1(𝑥+1)(𝑥3).

We want to consider this process in the reverse direction. That is to say, we want to split 3𝑥2(𝑥+1)(𝑥3) into two partial fractions: one with 𝑥+1 as the denominator and one with 𝑥3 as the denominator. Assuming we are working backward and have no knowledge of what the numerators are, we will have to assign them unknown values, 𝐴 and 𝐵. In other words, we have the equation 3𝑥1(𝑥+1)(𝑥3)=𝐴𝑥+1+𝐵𝑥3.

To calculate what 𝐴 and 𝐵 are, we can do the same process of adding the two fractions together as the one we did before, but this time with unknown constants: 𝐴𝑥+1+𝐵𝑥3=𝐴(𝑥3)(𝑥+1)(𝑥3)+𝐵(𝑥+1)(𝑥+1)(𝑥3)=𝐴(𝑥3)+𝐵(𝑥+1)(𝑥+1)(𝑥3).

Thus, 3𝑥1(𝑥+1)(𝑥3)=𝐴(𝑥3)+𝐵(𝑥+1)(𝑥+1)(𝑥3).

Since the denominators of the two algebraic fractions are equal, their numerators must also be equal. Therefore, we have 3𝑥1=𝐴(𝑥3)+𝐵(𝑥+1).

We can solve this equation in two ways. One way is to eliminate each factor on the right-hand side in turn by substituting values. For instance, by substituting 𝑥=3, we can eliminate the unknown 𝐴 to find 𝐵: 3(3)1=𝐴(33)+𝐵(3+1)8=0𝐴+4𝐵𝐵=2.

We can then substitute 𝑥=1 to eliminate the unknown 𝐵 and find 𝐴: 3(1)1=𝐴(13)+𝐵(1+1)4=4𝐴+0𝐵𝐴=1.

We can then substitute these values back into our original expression to obtain 3𝑥1(𝑥+1)(𝑥3)=1𝑥+1+2𝑥3.

We can see that this result indeed coincides with the result we got from performing the reverse process.

The second method we can use is to equate coefficients. We expand the brackets and collect like terms to get 3𝑥1=𝐴𝑥3𝐴+𝐵𝑥+𝐵3𝑥1=(𝐴+𝐵)𝑥3𝐴+𝐵.

The coefficients of 𝑥 must be equal on both sides of the equation. This gives us the equation 3=𝐴+𝐵.

Similarly, the constants on both sides of the equation must be equal: 1=3𝐴+𝐵.

Thus, we obtain a pair of linear simultaneous equations in two unknowns, which we can solve using the usual methods such as elimination or substitution.

There are a few cases where we cannot use this partial fraction decomposition.

Let us say we want to use partial fraction decomposition on 𝑃(𝑥)𝑄(𝑥), where 𝑃(𝑥) and 𝑄(𝑥) are polynomials and 𝑄(𝑥) is fully factored:

  • If the degree of 𝑃(𝑥) is greater than or equal to 𝑄(𝑥), then it is not possible to directly decompose 𝑃(𝑥)𝑄(𝑥) into partial fractions.
  • If 𝑃(𝑥) and 𝑄(𝑥) share a common factor, then we should first cancel the common factor before applying partial fraction decomposition.
  • If 𝑄(𝑥) has any repeated factors, then it is possible to use partial fraction decomposition; however, it is beyond the scope of this explainer to describe the method used.

In general, if the degree of 𝑃(𝑥) is less than the degree of 𝑄(𝑥) and 𝑄(𝑥) is a product of unique linear factors, then we can decompose 𝑃(𝑥)𝑄(𝑥) into the form 𝑃(𝑥)𝑄(𝑥)=𝐴𝑥𝑎++𝐴𝑥𝑎, where 𝑎,,𝑎 are all of the unique roots of 𝑄(𝑥) and 𝐴,,𝐴 are constants.

Usually, we will only apply partial fraction decomposition when there are two or three unique linear factors in the denominator. However, we can apply this method for any number of factors. We have the following.

Definition: Partial Fraction Decomposition

If the degree of 𝑃(𝑥) is less than 2 and 𝑄(𝑥)=(𝑥𝑎)(𝑥𝑏) is a product of unique linear factors, then we can decompose 𝑃(𝑥)𝑄(𝑥) into the form 𝑃(𝑥)𝑄(𝑥)=𝐴𝑥𝑎+𝐵𝑥𝑏, for some unknown constants 𝐴 and 𝐵.

Similarly, if the degree of 𝑃(𝑥) is less than 3 and 𝑄(𝑥)=(𝑥𝑎)(𝑥𝑏)(𝑥𝑐) is a product of unique linear factors, then we can decompose 𝑃(𝑥)𝑄(𝑥) into the form 𝑃(𝑥)𝑄(𝑥)=𝐴𝑥𝑎+𝐵𝑥𝑏+𝐶𝑥𝑐, for some unknown constants 𝐴, 𝐵, and 𝐶.

Let us consider an example of this process.

Example 1: Finding Unknowns in Partial Fractions with Two Nonrepeated Linear Factors

The expression 2𝑥+1(𝑥+2)(𝑥+3) can be written in the form 𝐴𝑥+3+𝐵𝑥+2. Find the values of 𝐴 and 𝐵.

Answer

We are told that 2𝑥+1(𝑥+2)(𝑥+3)=𝐴𝑥+3+𝐵𝑥+2, for some unknown values 𝐴 and 𝐵. To find these values, we can add the two partial fractions on the right-hand side together.

We recall that to add algebraic fractions together, we first need to make sure that they have the same denominator, and we can do this by cross multiplying: 𝐴𝑥+3+𝐵𝑥+2=𝐴(𝑥+2)(𝑥+3)(𝑥+2)+𝐵(𝑥+3)(𝑥+3)(𝑥+2)=𝐴(𝑥+2)+𝐵(𝑥+3)(𝑥+3)(𝑥+2).

This expression is equal to the left-hand side of the original equation we are given, so 2𝑥+1(𝑥+2)(𝑥+3)=𝐴(𝑥+2)+𝐵(𝑥+3)(𝑥+3)(𝑥+2).

We now note that the denominators are the same on both sides of the equation. Hence, the numerators must be the same: 2𝑥+1=𝐴(𝑥+2)+𝐵(𝑥+3).

There are now two methods we can use to find 𝐴 and 𝐵 and it is a personal preference which one we may want to use.

The first method involves substitution. We can substitute 𝑥=2 into the equation, where we note that this will remove 𝐴 from the equation: 2(2)+1=𝐴(2+2)+𝐵(2+3)3=0𝐴+𝐵𝐵=3.

We can substitute 𝑥=3 into the equation to eliminate the unknown 𝐵 from the equation: 2(3)+1=𝐴(3+2)+𝐵(3+3)5=𝐴+0𝐵𝐴=5.

The second method we can use involves expanding the brackets and collecting like terms to obtain 2𝑥+1=𝐴(𝑥+2)+𝐵(𝑥+3)2𝑥+1=𝐴𝑥+2𝐴+𝐵𝑥+3𝐵2𝑥+1=(𝐴+𝐵)𝑥+2𝐴+3𝐵.

We can now equate the coefficients to construct equations involving 𝐴 and 𝐵. Equating the coefficients of 𝑥 yields 2=𝐴+𝐵.

Equating the constants then gives us 1=2𝐴+3𝐵.

We can solve these two equations as simultaneous equations. Rearranging the first equation gives us 𝐴=2𝐵.

Substituting this into the second equation and solving gives us 1=2(2𝐵)+3𝐵1=42𝐵+3𝐵1=4+𝐵𝐵=3.

We can then substitute 𝐵=3 into the equation 𝐴=2𝐵 to find 𝐴=2(3)=5.

Hence, 𝐴=5 and 𝐵=3.

In our next example, we will consider a partial fraction decomposition where we are not given the form the partial fractions take.

Example 2: Partial Fraction Decomposition with Two Nonrepeated Linear Factors

Express 𝑥2𝑥(𝑥3) in partial fractions.

Answer

We first note that we have nonrepeated linear factors in the denominator and the degree of the polynomial in the numerator is less than that of 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 as denominators: 𝑥2𝑥(𝑥3)=𝐴𝑥+𝐵𝑥3, for some unknown constants 𝐴 and 𝐵.

We can solve this equation for 𝐴 and 𝐵 by adding the fractions on the right-hand side of the equation together. We first rewrite the fractions to have the same denominator by cross multiplying: 𝐴𝑥+𝐵𝑥3=𝐴(𝑥3)𝑥(𝑥3)+𝐵𝑥𝑥(𝑥3).

Since the fractions have the same denominator, we can now add the numerators to obtain 𝐴(𝑥3)𝑥(𝑥3)+𝐵𝑥𝑥(𝑥3)=𝐴(𝑥3)+𝐵𝑥𝑥(𝑥3).

This expression needs to be equal to the original expression we are given in the question. Thus, 𝑥2𝑥(𝑥3)=𝐴(𝑥3)+𝐵𝑥𝑥(𝑥3)

We now note that the denominators are the same on both sides of the equation. Hence, the numerators must be the same: 𝑥2=𝐴(𝑥3)+𝐵𝑥.

We can find the values of 𝐴 and 𝐵 by using substitution. Substituting 𝑥=3 into the equation and solving yields 32=𝐴(33)+𝐵(3)1=3𝐵𝐵=13.

Substituting 𝑥=0 into the equation and solving yields 02=𝐴(03)+𝐵(0)2=3𝐴𝐴=23.

We can now substitute 𝐴=23 and 𝐵=13 into the partial fraction decomposition and simplify to obtain 𝑥2𝑥(𝑥3)=𝐴𝑥+𝐵𝑥3=𝑥+𝑥3=23𝑥+13(𝑥3).

Hence, 𝑥2𝑥(𝑥3)=23𝑥+13(𝑥3).

So far, we have only investigated cases of splitting an algebraic fraction into two partial fractions. In the next example, we will consider the case where the denominator is a cubic polynomial that will be decomposed into three partial fractions. However, as we will see, this changes very little about the method, other than it requiring finding more constants and multiplication by more factors.

Example 3: Partial Fraction Decomposition with Three Unique Linear Factors

Express 𝑥2(𝑥+2)(𝑥3)(𝑥+1) in partial fractions.

Answer

We first note that we have nonrepeated linear factors in the denominator, and the degree of the polynomial in the numerator is less than that of 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 as denominators: 𝑥2(𝑥+2)(𝑥3)(𝑥+1)=𝐴𝑥+1+𝐵𝑥+2+𝐶𝑥3, for some constants 𝐴, 𝐵, and 𝐶.

To determine the values of 𝐴, 𝐵, and 𝐶, we need to add the algebraic fractions on the right-hand side of the equation together. To do this, we need their denominators to be equal, so we multiply the numerator and denominator of each fraction to make the shared denominator the product of the linear factors: 𝐴𝑥+1+𝐵𝑥+2+𝐶𝑥3=𝐴(𝑥+2)(𝑥3)(𝑥+1)(𝑥+2)(𝑥3)+𝐵(𝑥3)(𝑥+1)(𝑥+2)(𝑥3)(𝑥+1)+𝐶(𝑥+2)(𝑥+1)(𝑥3)(𝑥+2)(𝑥+1).

Now that the denominators are equal, we can add the fractions together by adding their numerators to obtain 𝐴(𝑥+2)(𝑥3)+𝐵(𝑥3)(𝑥+1)+𝐶(𝑥+2)(𝑥+1)(𝑥+2)(𝑥3)(𝑥+1).

This expression must be equal to the expression given to us in the question. Hence, we have 𝑥2(𝑥+2)(𝑥3)(𝑥+1)=𝐴(𝑥+2)(𝑥3)+𝐵(𝑥3)(𝑥+1)+𝐶(𝑥+2)(𝑥+1)(𝑥+2)(𝑥3)(𝑥+1).

Since the denominators of both sides of the equation are equal, the numerators must be equal: 𝑥2=𝐴(𝑥+2)(𝑥3)+𝐵(𝑥3)(𝑥+1)+𝐶(𝑥+2)(𝑥+1).

We can now determine the values of 𝐴, 𝐵, and 𝐶 by substitution. First, we substitute 𝑥=3 into the equation to get 32=𝐴(3+2)(33)+𝐵(33)(3+1)+𝐶(3+2)(3+1)7=20𝐶𝐶=720.

Next, we substitute 𝑥=2 into the equation to obtain (2)2=𝐴(2+2)(23)+𝐵(23)(2+1)+𝐶(2+2)(2+1)2=5𝐵𝐵=25.

Finaly, we substitute 𝑥=1 into the equation, giving us (1)2=𝐴(1+2)(13)+𝐵(13)(1+1)+𝐶(1+2)(1+1)1=4𝐴𝐴=14.

We can now substitute these values into the partial fraction decomposition and simplify to find 𝑥2(𝑥+2)(𝑥3)(𝑥+1)=𝐴𝑥+1+𝐵𝑥+2+𝐶𝑥3=𝑥+1+𝑥+2+𝑥3=14(𝑥+1)+25(𝑥+2)+720(𝑥3).

Hence, 𝑥2(𝑥+2)(𝑥3)(𝑥+1)=14(𝑥+1)+25(𝑥+2)+720(𝑥3).

It is worth noting that when we are familiar with the process of partial fraction decomposition, it is not necessary to write out all of the steps. For instance, in the previous example, once we have the equation 𝑥2(𝑥+2)(𝑥3)(𝑥+1)=𝐴(𝑥+2)(𝑥3)+𝐵(𝑥3)(𝑥+1)+𝐶(𝑥+2)(𝑥+1)(𝑥+2)(𝑥3)(𝑥+1). we could substitute values of 𝑥 directly into the numerators or equate coefficients in the numerators without writing an equation explicitly equating the numerators.

In our next example, we will consider something slightly different. Instead of being given the numerator of the fraction we want to decompose, we will be given its value at certain points and we will have to use this information to perform the partial fraction decomposition.

Example 4: Setting Up and Applying Partial Fractions with Three Nonrepeated Linear Factors

Decompose 𝑃(𝑥)(𝑥1)(𝑥3)(𝑥5) into partial fractions, where 𝑃(𝑥) is a quadratic function, 𝑃(1)=8, 𝑃(3)=8, and 𝑃(5)=24.

Answer

We might be tempted to try and determine the polynomial 𝑃(𝑥); however, it is not necessary to explicitly find it since we know it is a quadratic and we have 𝑃(𝑥) evaluated at three values of 𝑥.

We now note that we have nonrepeated linear factors in the denominator and the degree of the polynomial in the numerator is less than that of the denominator. We can recall that this means that we can use partial fractions to rewrite this expression in terms of fractions, with the linear factors as denominators: 𝑃(𝑥)(𝑥1)(𝑥3)(𝑥5)=𝐴𝑥1+𝐵𝑥3+𝐶𝑥5, for some constants 𝐴, 𝐵, and 𝐶.

We can determine the constants by adding the three algebraic fractions together by rewriting them to have the same denominators. We have 𝐴𝑥1+𝐵𝑥3+𝐶𝑥5=𝐴(𝑥3)(𝑥5)(𝑥1)(𝑥3)(𝑥5)+𝐵(𝑥1)(𝑥5)(𝑥1)(𝑥3)(𝑥5)+𝐶(𝑥1)(𝑥3)(𝑥1)(𝑥3)(𝑥5).

Now that the algebraic fractions all have the same denominator, we can add them together by adding their numerators: 𝑃(𝑥)(𝑥1)(𝑥3)(𝑥5)=𝐴(𝑥3)(𝑥5)+𝐵(𝑥1)(𝑥5)+𝐶(𝑥1)(𝑥3)(𝑥1)(𝑥3)(𝑥5).

Since the denominators on both sides of the equation are equal, the numerators must also be equal: 𝑃(𝑥)=𝐴(𝑥3)(𝑥5)+𝐵(𝑥1)(𝑥5)+𝐶(𝑥1)(𝑥3).

We can now substitute the values of 𝑥 into the equation that we are given in the question.

First, we are told that 𝑃(1)=8, so when 𝑥=1, we have 8=𝐴(13)(15)+𝐵(11)(𝑥5)+𝐶(11)(𝑥3)8=8𝐴𝐴=1.

Second, we are told that 𝑃(3)=8, so when 𝑥=3, we have 8=𝐴(33)(35)+𝐵(31)(35)+𝐶(31)(33)8=4𝐵𝐵=2. Finally, we are told that 𝑃(5)=24, so when 𝑥=5, we have 24=𝐴(53)(55)+𝐵(51)(55)+𝐶(51)(53)24=8𝐶𝐶=3.

Hence, 𝑃(𝑥)(𝑥1)(𝑥3)(𝑥5)=1𝑥1+2𝑥3+3𝑥5.

In our next example, we will decompose an algebraic fraction into partial fractions by first factorizing its denominator.

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

Express 𝑥23𝑥𝑥20 in partial fractions.

Answer

In order to apply partial fraction decomposition, we first need the denominator of the algebraic fraction to be fully factored. We can factor this monic quadratic by finding two numbers whose product is 20 and whose sum is 4. By considering the factors of 20, we find that these are 5 and 4. So, 𝑥𝑥20=(𝑥+4)(𝑥5) and 𝑥23𝑥𝑥20=𝑥23(𝑥+4)(𝑥5).

We now note that we have nonrepeated linear factors in the denominator and the degree of the polynomial in the numerator is less than that of the denominator. We can recall that this means that we can use partial fractions to rewrite the expression in terms of fractions, with the linear factors as denominators: 𝑥23(𝑥+4)(𝑥5)=𝐴𝑥+4+𝐵𝑥5,

for some unknown values of 𝐴 and 𝐵.

We can determine the values of 𝐴 and 𝐵 by adding the algebraic fractions on the right-hand side of the equation together. We add them by rewriting them to have the same denominator: 𝐴𝑥+4+𝐵𝑥5=𝐴(𝑥5)(𝑥+4)(𝑥5)+𝐵(𝑥+4)(𝑥5)(𝑥+4)=𝐴(𝑥5)+𝐵(𝑥+4)(𝑥+4)(𝑥5)

This must be equal to the algebraic fraction we want to decompose: 𝑥23(𝑥+4)(𝑥5)=𝐴(𝑥5)+𝐵(𝑥+4)(𝑥+4)(𝑥5).

We now note that the denominators on both sides of the equation are equal, so their numerators must also be equal: 𝑥23=𝐴(𝑥5)+𝐵(𝑥+4).

We can solve for 𝐴 and 𝐵 by substitution. When 𝑥=5, we have 523=𝐴(55)+𝐵(5+4)18=9𝐵𝐵=2.

When 𝑥=4, we find 423=𝐴(45)+𝐵(4+4)27=9𝐴𝐴=3.

Hence, 𝑥23(𝑥+4)(𝑥5)=3𝑥+42𝑥5.

In our final example, let us consider an algebraic fraction with an unfactored quadratic and a cubic in the numerator and denominator respectively.

Example 6: Factoring Denominators and Applying Partial Fractions with Three Nonrepeated Linear Factors

Express 2𝑥9𝑥+1𝑥𝑥2𝑥 in partial fractions.

Answer

In order to apply partial fraction decomposition, we first need the denominator of the algebraic fraction to be fully factorized. We can factorize the denominator by noting that each term shares a factor of 𝑥: 𝑥𝑥2𝑥=𝑥𝑥𝑥2.

We can then factor the monic quadratic by finding two numbers whose product is 2 and whose sum is 1: 𝑥𝑥𝑥2=𝑥(𝑥+1)(𝑥2).

Thus, 2𝑥9𝑥+1𝑥𝑥2𝑥=2𝑥9𝑥+1𝑥(𝑥+1)(𝑥2).

We now note that we have nonrepeated linear factors in the denominator, and the degree of the polynomial in the numerator is less than that of the denominator. We can recall that this means that we can use partial fractions to rewrite the expression in terms of fractions, with the linear factors as denominators: 2𝑥9𝑥+1𝑥(𝑥+1)(𝑥2)=𝐴𝑥+𝐵𝑥+1+𝐶𝑥2, for some unknown values of 𝐴, 𝐵, and 𝐶.

We can determine the values of these constants by adding the algebraic fractions on the right-hand side of the equation together. We add them by rewriting them to have the same denominator: 𝐴𝑥+𝐵𝑥+1+𝐶𝑥2=𝐴(𝑥+1)(𝑥2)+𝐵𝑥(𝑥2)+𝐶𝑥(𝑥+1)𝑥(𝑥+1)(𝑥2).

This must be equal to the algebraic fraction we want to decompose: 2𝑥9𝑥+1𝑥(𝑥+1)(𝑥2)=𝐴(𝑥+1)(𝑥2)+𝐵𝑥(𝑥2)+𝐶𝑥(𝑥+1)𝑥(𝑥+1)(𝑥2).

We now note that the denominators on both sides of the equation are equal, so their numerators must also be equal: 2𝑥9𝑥+1=𝐴(𝑥+1)(𝑥2)+𝐵𝑥(𝑥2)+𝐶𝑥(𝑥+1).

We can solve for 𝐴, 𝐵, and 𝐶 by substitution. When 𝑥=0, we have 2(0)9(0)+1=𝐴(0+1)(02)+𝐵(0)(02)+𝐶(0)(0+1)1=2𝐴𝐴=12.

When 𝑥=1, we have 2(1)9(1)+1=𝐴(1+1)(12)+𝐵(1)(12)+𝐶(1)(1+1)12=3𝐵𝐵=4.

When 𝑥=2, we have 2(2)9(2)+1=𝐴(2+1)(22)+𝐵(2)(22)+𝐶(2)(2+1)9=6𝐶𝐶=32.

Hence, 2𝑥9𝑥+1𝑥(𝑥+1)(𝑥2)=𝐴𝑥+𝐵𝑥+1+𝐶𝑥2=𝑥+4𝑥+1+𝑥2=12𝑥+4𝑥+132(𝑥2).

Let us finish by recapping some of the important points from this explainer.

Key Points

  • If the degree of 𝑃(𝑥) is less than 2 and 𝑄(𝑥)=(𝑥𝑎)(𝑥𝑏) is a product of unique linear factors, then we can decompose 𝑃(𝑥)𝑄(𝑥) into partial fractions of the form 𝑃(𝑥)𝑄(𝑥)=𝐴𝑥𝑎+𝐵𝑥𝑏, for some unknown constants 𝐴 and 𝐵.
  • If the degree of 𝑃(𝑥) is less than 3 and 𝑄(𝑥)=(𝑥𝑎)(𝑥𝑏)(𝑥𝑐) is a product of unique linear factors, then we can decompose 𝑃(𝑥)𝑄(𝑥) into partial fractions of the form 𝑃(𝑥)𝑄(𝑥)=𝐴𝑥𝑎+𝐵𝑥𝑏+𝐶𝑥𝑐, for some unknown constants 𝐴, 𝐵, and 𝐶.
  • There are two methods we can use to find the unknown constants: substitution and equating coefficients. It is important to be comfortable with both methods.
  • 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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