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.

We find that in this case, if 𝑄(𝑥)=𝑘(𝑥𝑎)(𝑥𝑎)(𝑥𝑎) with 𝑛 different numbers 𝑎,𝑎,,𝑎 together with (leading coefficient) 𝑘, and if we also assume that the degree of 𝑃(𝑥) is less than 𝑛, the degree of 𝑄(𝑥), then 𝑃(𝑥)𝑄(𝑥)=𝑃(𝑥)𝑘(𝑥𝑎)(𝑥𝑎)(𝑥𝑎)=𝐴(𝑥𝑎)+𝐴(𝑥𝑎)++𝐴(𝑥𝑎) for numbers 𝐴,𝐴,,𝐴, which can be found by solving a set of linear equations.

For example, consider 5𝑥3𝑥322𝑥8𝑥+2𝑥+12=5𝑥3𝑥322(𝑥+1)(𝑥2)(𝑥3). The denominator 𝑄(𝑥)=2(𝑥+1)(𝑥2)(𝑥3) has the form that we are interested in. We expect 5𝑥3𝑥322(𝑥+1)(𝑥2)(𝑥3)=𝐴𝑥+1+𝐵𝑥2+𝐶𝑥3𝐴,𝐵,𝐶.forsomenumbers

The simplest way to find these is to multiply this identity through with 𝑄(𝑥), or 2(𝑥+1)(𝑥2)(𝑥3). This gives the identity 5𝑥3𝑥32=2𝐴(𝑥2)(𝑥3)+2𝐵(𝑥+1)(𝑥3)+2𝐶(𝑥+1)(𝑥2) by the way the factors cancel each other out.

To determine these constants, we consider what happens if we substitute 𝑥=1 in this equation. It produces the identity 5(1)3(1)32=2𝐴(12)(13)+2𝐵(1+1)(13)+2𝐶(1+1)(12)5+332=24𝐴+2𝐵(0)+2𝐶(0)24=24𝐴 such that 𝐴=1. What made this work was the fact that the other two terms on the right-hand side vanished at 𝑥=1. Likewise, looking at 𝑥=2 and 𝑥=3 respectively gives us 5(2)3(2)32=18=2𝐵(3)5(3)3(3)32=4=2𝐶(4) such that 𝐵=3 and 𝐶=12. We have found the partial fraction decomposition: 5𝑥3𝑥322(𝑥+1)(𝑥2)(𝑥3)=1𝑥+1+3𝑥2+12(𝑥3). What happens if degdeg(𝑃(𝑥))(𝑄(𝑥))? In this case, we start with the long division of polynomials and write 𝑃(𝑥)𝑄(𝑥)=𝑆(𝑥)+𝑃(𝑥)𝑄(𝑥) with quotient polynomial 𝑆(𝑥) and a remainder polynomial 𝑃(𝑥) of smaller degree than 𝑄(𝑥), such that the above method now applies to 𝑃(𝑥)𝑄(𝑥).

We summarize as in the following.

Partial Fractions: When 𝑄(𝑥) Splits into Linear Factors, None Repeated

  1. If 𝑃(𝑥) has a degree greater than or equal to the degree of 𝑄(𝑥), apply the division algorithm to take a polynomial quotient from it. After this, assume degdeg(𝑃(𝑥))<(𝑄(𝑥)).
  2. If 𝑄(𝑥) has 𝑛 factors (𝑥𝑎),(𝑥𝑎),,(𝑥𝑎), then there are numbers 𝐴,,𝐴 such that the partial fraction decomposition is 𝑃(𝑥)𝑄(𝑥)=𝐴(𝑥𝑎)+𝐴(𝑥𝑎)++𝐴(𝑥𝑎).
  3. To find 𝐴, multiply through by 𝑄(𝑥) to get the equation 𝑃(𝑥)=𝐴𝑞(𝑥)+𝐴𝑞(𝑥)++𝐴𝑞(𝑥) with each of the 𝑞(𝑥) a product of the factors (including the constant one) of 𝑄(𝑥) other than (𝑥𝑎).
  4. Substituting 𝑥=𝑎, we find that 𝑞(𝑎)0 but 𝑞(𝑎)=𝑞(𝑎)=𝑞(𝑎)=0, which gives the relation 𝑃(𝑎)=𝐴𝑞(𝑎) that we solve for 𝐴.
  5. Repeat the steps above to find 𝐴,𝐴,,𝐴.

Example 1: Decomposing Rational Expressions into Partial Fractions

Find 𝐴 and 𝐵 such that 4𝑥2(𝑥+3)(𝑥2)=𝐴𝑥+3+𝐵𝑥2.


Here, the degree of 𝑃(𝑥)=4𝑥2 is less than the degree of 𝑄(𝑥)=(𝑥+3)(𝑥2). Multiplying through by this 𝑄(𝑥) gives the equation 4𝑥2=𝐴(𝑥2)+𝐵(𝑥+3). Setting 𝑥=3 gives 4(3)2=𝐴(32)14=5𝐴𝐴=145. Setting 𝑥=2 gives 4(2)2=𝐵(2+3)6=5𝐵𝐵=65.

Our solution is therefore 𝐴=145,𝐵=65.

Example 2: Decomposing Rational Expressions into Partial Fractions

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


The denominator has a larger degree than the numerator, so we proceed to find 𝐴,𝐵,𝐶 such that 𝑥2(𝑥+2)(𝑥3)(𝑥+1)=𝐴𝑥+2+𝐵𝑥3+𝐶𝑥+1.

Multiplying through by (𝑥+2)(𝑥3)(𝑥+1), we get 𝑥2=𝐴(𝑥3)(𝑥+1)+𝐵(𝑥+2)(𝑥+1)+𝐶(𝑥+2)(𝑥3).

Then substituting 𝑥=2,3, and 1 gives equations 2=𝐴(23)(2+1)=5𝐴,7=𝐵(3+2)(3+1)=20𝐵,1=𝐶(1+2)(13)=4𝐶 such that 𝐴=15,𝐵=720,𝐶=14.

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