Explainer: Partial Fractions: Nonrepeated Linear Factors

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(𝑥𝑎1)(𝑥𝑎2)(𝑥𝑎𝑛) with 𝑛 different numbers 𝑎1,𝑎2,,𝑎𝑛 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(𝑥𝑎1)(𝑥𝑎2)(𝑥𝑎𝑛)=𝐴1(𝑥𝑎1)+𝐴2(𝑥𝑎2)++𝐴𝑛(𝑥𝑎𝑛) for numbers 𝐴1,𝐴2,,𝐴𝑛, which can be found by solving a set of linear equations.

For example, consider 5𝑥23𝑥322𝑥38𝑥2+2𝑥+12=5𝑥23𝑥322(𝑥+1)(𝑥2)(𝑥3). The denominator 𝑄(𝑥)=2(𝑥+1)(𝑥2)(𝑥3) has the form that we are interested in. We expect 5𝑥23𝑥322(𝑥+1)(𝑥2)(𝑥3)=𝐴𝑥+1+𝐵𝑥2+𝐶𝑥3forsomenumbers𝐴,𝐵,𝐶.

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 5𝑥23𝑥32=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)23(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)23(2)32=18=2𝐵(3)5(3)23(3)32=4=2𝐶(4) such that 𝐵=3 and 𝐶=12. We have found the partial fraction decomposition: 5𝑥23𝑥322(𝑥+1)(𝑥2)(𝑥3)=1𝑥+1+3𝑥2+12(𝑥3). What happens if deg(𝑃(𝑥))deg(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 deg(𝑃(𝑥))<deg(𝑄(𝑥)).
  2. If 𝑄(𝑥) has 𝑛 factors (𝑥𝑎1),(𝑥𝑎2),,(𝑥𝑎𝑛), then there are numbers 𝐴1,,𝐴𝑛 such that the partial fraction decomposition is 𝑃(𝑥)𝑄(𝑥)=𝐴1(𝑥𝑎1)+𝐴2(𝑥𝑎2)++𝐴𝑛(𝑥𝑎𝑛).
  3. To find 𝐴1, multiply through by 𝑄(𝑥) to get the equation 𝑃(𝑥)=𝐴1𝑞1(𝑥)+𝐴2𝑞2(𝑥)++𝐴𝑛𝑞𝑛(𝑥) with each of the 𝑞𝑖(𝑥) a product of the factors (including the constant one) of 𝑄(𝑥) other than (𝑥𝑎𝑖).
  4. Substituting 𝑥=𝑎1, we find that 𝑞1(𝑎1)0 but 𝑞2(𝑎1)=𝑞3(𝑎1)=𝑞𝑛(𝑎1)=0, which gives the relation 𝑃(𝑎1)=𝐴1𝑞1(𝑎1) that we solve for 𝐴1.
  5. Repeat the steps above to find 𝐴2,𝐴3,,𝐴𝑛.

Example 1: Decomposing Rational Expressions into Partial Fractions

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

Answer

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 𝑥22(𝑥+2)(𝑥3)(𝑥+1) in partial fractions.

Answer

The denominator has a larger degree than the numerator, so we proceed to find 𝐴,𝐵,𝐶 such that 𝑥22(𝑥+2)(𝑥3)(𝑥+1)=𝐴𝑥+2+𝐵𝑥3+𝐶𝑥+1.

Multiplying through by (𝑥+2)(𝑥3)(𝑥+1), we get 𝑥22=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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