Explainer: Partial Fractions: Repeated Linear Factors

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

For example, consider 𝑃(π‘₯)𝑄(π‘₯)=1+π‘₯+π‘₯(π‘₯βˆ’1)(π‘₯βˆ’2)(π‘₯βˆ’3).

When expanded, we see that 𝑄(π‘₯) has degree 6=1+2+3, while the degree of 𝑃(π‘₯) is less at 5. The result is that in this case, we can determine a partial fraction expansion like this: 𝐴π‘₯βˆ’1+𝐡π‘₯βˆ’2+𝐢(π‘₯βˆ’2)+𝐷π‘₯βˆ’3+𝐸(π‘₯βˆ’3)+𝐹(π‘₯βˆ’3).

Structure of Partial Fraction Expansion: Repeated Linear Factors

The partial fraction expansion is a sum of terms, where

  1. each term has the form π‘˜(π‘₯βˆ’π‘Ž),
  2. π‘˜ is a constant,
  3. π‘Ž is one of the roots of 𝑄(π‘₯), so corresponds to a distinct linear factor, and
  4. the exponent π‘š satisfies π‘šβ‰€π‘›, where (π‘₯βˆ’π‘Ž) is the factor of 𝑄(π‘₯) corresponding to π‘Ž.

To see why this must be the case, consider 𝑃(π‘₯)𝑄(π‘₯)=𝑃(π‘₯)(π‘₯βˆ’π‘Ž)(π‘₯βˆ’π‘),π‘Ž,π‘βˆˆβ„οŠ¨ with the degree of 𝑃(π‘₯) at most 2. Multiplying through by (π‘₯βˆ’π‘Ž), (π‘₯βˆ’π‘Ž)𝑃(π‘₯)𝑄(π‘₯)=𝑃(π‘₯)(π‘₯βˆ’π‘Ž)(π‘₯βˆ’π‘)=𝐴(π‘₯βˆ’π‘Ž)+𝐡(π‘₯βˆ’π‘)+𝐢 for some constants 𝐴, 𝐡, and 𝐢.

The first two terms are what you expect of the partial fraction expansion of 𝑃(π‘₯)(π‘₯βˆ’π‘Ž)(π‘₯βˆ’π‘) in case the degree of 𝑃(π‘₯) is at most 1. Here, the degree may be 2, in which case long division gives a nonzero constant 𝐢. Dividing both sides by (π‘₯βˆ’π‘Ž) gives 𝑃(π‘₯)𝑄(π‘₯)=𝐴(π‘₯βˆ’π‘Ž)+𝐡(π‘₯βˆ’π‘Ž)(π‘₯βˆ’π‘)+𝐢(π‘₯βˆ’π‘Ž) and applying the partial fraction expansion to the middle term 𝐴(π‘₯βˆ’π‘Ž)+𝐡(π‘₯βˆ’π‘Ž)(π‘₯βˆ’π‘)+𝐢(π‘₯βˆ’π‘Ž)=𝐴(π‘₯βˆ’π‘Ž)+𝐸(π‘₯βˆ’π‘Ž)+𝐹(π‘₯βˆ’π‘)+𝐢(π‘₯βˆ’π‘Ž).

If we set 𝐺=𝐸+𝐢, on gathering the two like terms, we get 𝐴(π‘₯βˆ’π‘Ž)+𝐺(π‘₯βˆ’π‘Ž)+𝐹(π‘₯βˆ’π‘).

Extending this argument, we see that any rational expression of the form 𝑃(π‘₯)(π‘₯βˆ’π‘Ž)(π‘₯βˆ’π‘)𝑃(π‘₯)(π‘₯βˆ’π‘Ž)(π‘₯βˆ’π‘),(𝑃(π‘₯))≀3,orwheredeg has the right kind of partial fraction expansion.

Here is a list of steps to follow to determine a partial fraction expansion, illustrated with this example: 3π‘₯βˆ’13π‘₯+14(π‘₯βˆ’3)(π‘₯βˆ’1).

  1. If necessary, factorize the denominator (already done in this example). We note the form we expect: 𝐴(π‘₯βˆ’3)+𝐡(π‘₯βˆ’3)+𝐢(π‘₯βˆ’1) with unknown constants 𝐴, 𝐡, and 𝐢. This is because the degree of the numerator here was less than 3. Otherwise, we would have begun with a long division.
  2. Multiplying this form with the denominator (π‘₯βˆ’3)(π‘₯βˆ’1) should give us the numerator 3π‘₯βˆ’13π‘₯+14. This is the equation 3π‘₯βˆ’13π‘₯+14=𝐴(π‘₯βˆ’1)+𝐡(π‘₯βˆ’1)(π‘₯βˆ’3)+𝐢(π‘₯βˆ’3).
  3. We solve for these constants by substituting values of π‘₯ that allow us to β€œpick off” values. For example, setting π‘₯=1 gives us 3(1)βˆ’13(1)+14=𝐴(1βˆ’1)+𝐡(1βˆ’1)(1βˆ’3)+𝐢(1βˆ’3)4=𝐢(βˆ’2)∴𝐢=1.
    Setting π‘₯=3 gives us
    3(3)βˆ’13(3)+14=𝐴(3βˆ’1)+𝐡(3βˆ’1)(3βˆ’3)+𝐢(3βˆ’3)2=𝐴(2)∴𝐴=1.

For the middle term, we substitute any value we like, keeping in mind that we now know 𝐴 and 𝐢. Setting π‘₯=2, for example, and using the found values: 3(2)βˆ’13(2)+14=(1)(2βˆ’1)+𝐡(2βˆ’1)(2βˆ’3)+(1)(2βˆ’3)0=1+𝐡(1)(βˆ’1)+(1)(βˆ’1)0=2βˆ’π΅βˆ΄π΅=2.

Hence, the partial fraction decomposition 3π‘₯βˆ’13π‘₯+14(π‘₯βˆ’3)(π‘₯βˆ’1)=1(π‘₯βˆ’3)+2(π‘₯βˆ’3)+1(π‘₯βˆ’1).

Example 1: Decomposing Rational Expressions into Partial Fractions

Find 𝐴 and 𝐡 such that 2π‘₯(π‘₯βˆ’3)=𝐴π‘₯βˆ’3+𝐡(π‘₯βˆ’3).

Answer

Here, the degree of 𝑃(π‘₯)=2π‘₯ is less than the degree of 𝑄(π‘₯)=ο€Ήπ‘₯βˆ’3ο…οŠ¨. Multiplying through by this 𝑄(π‘₯) gives the equation 2π‘₯=𝐴(π‘₯βˆ’3)+𝐡.

Setting π‘₯=3 gives 2(3)=𝐴(3βˆ’3)+𝐡6=𝐡𝐡=6.

We could set 𝐡 to its value and use an arbitrary value for π‘₯ and solve for 𝐴 in a second equation. But easier is to note that the equation above is of polynomials in π‘₯. The leading coefficient of the right-hand side must be 2, so 𝐴=2. Our solution is therefore 𝐴=2,𝐡=6.

Suppose that we do not only have linear factors in the denominator. What can we expect?

The answer is that we treat irreducible factors of 𝑄(π‘₯) as if they were linear terms, except that instead of constants, we seek polynomials of lower degree than the factor, so π‘₯+4π‘₯βˆ’2(π‘₯βˆ’1)(π‘₯+2)=𝐴(π‘₯βˆ’1)+𝐡(π‘₯βˆ’1)ο‡Œο†²ο†²ο†²ο†²ο‡ο†²ο†²ο†²ο†²ο‡Ž+𝐢π‘₯+𝐷(π‘₯+2) for some constants 𝐴, 𝐡, 𝐢, and 𝐷 that must be determined. The first two terms are as above for repeated linear factors. The last has a degree 1 numerator over the irreducible factor ο€Ήπ‘₯+2ο…οŠ¨.

Here is a systematic way to solve this example. First, we write this as an equality of numerators, by multiplying the right-hand side by (π‘₯βˆ’1)(π‘₯+2): π‘₯+4π‘₯βˆ’2=𝐴π‘₯+2+𝐡π‘₯+2(π‘₯βˆ’1)+(𝐢π‘₯+𝐷)(π‘₯βˆ’1) and then substituting π‘₯=1 on both sides allows to solve for 𝐴, since 1+4ο€Ή1ο…βˆ’2=𝐴1+23=3𝐴∴𝐴=1.

Next, we rewrite this, substituting 𝐴=1 and maintaining the unexpanded terms, as an equation set to zero: ο€Ήπ‘₯+2+𝐡π‘₯+2(π‘₯βˆ’1)+(𝐢π‘₯+𝐷)(π‘₯βˆ’1)=π‘₯+4π‘₯βˆ’2ο€Ήπ‘₯+2+𝐡π‘₯+2(π‘₯βˆ’1)+(𝐢π‘₯+𝐷)(π‘₯βˆ’1)βˆ’3=ο€Ήπ‘₯+4π‘₯βˆ’2ο…βˆ’3ο€Ήπ‘₯βˆ’1+𝐡π‘₯+2(π‘₯βˆ’1)+(𝐢π‘₯+𝐷)(π‘₯βˆ’1)=π‘₯+4π‘₯βˆ’5, where we have subtracted the value 3 at π‘₯=1 from both sides. This means that both sides are now equal to zero at π‘₯=1, and so contain a factor (π‘₯βˆ’1), which we cancel: ο€Ήπ‘₯βˆ’1+𝐡π‘₯+2(π‘₯βˆ’1)+(𝐢π‘₯+𝐷)(π‘₯βˆ’1)=(π‘₯βˆ’1)ο€Ήπ‘₯+5π‘₯+5(π‘₯βˆ’1)(π‘₯+1)+𝐡π‘₯+2+(𝐢π‘₯+𝐷)(π‘₯βˆ’1)=(π‘₯βˆ’1)ο€Ήπ‘₯+5π‘₯+5(π‘₯+1)+𝐡π‘₯+2+(𝐢π‘₯+𝐷)(π‘₯βˆ’1)=π‘₯+5π‘₯+5.

The reason this helps is that now we can substitute π‘₯=1 once more to determine 𝐡: (1+1)+𝐡1+2+(𝐢(1)+2𝐷)(1βˆ’1)=1+5(1)+53𝐡+2=11∴𝐡=3.

Finally, we can solve for 𝐢 and 𝐷 by setting π‘₯=0 in the original equation (with updated values of 𝐴 and 𝐡) to obtain just 𝐷: π‘₯+4π‘₯βˆ’2=𝐴π‘₯+2+𝐡π‘₯+2(π‘₯βˆ’1)+(𝐢π‘₯+𝐷)(π‘₯βˆ’1)=ο€Ήπ‘₯+2+3ο€Ήπ‘₯+2(π‘₯βˆ’1)+(𝐢π‘₯+𝐷)(π‘₯βˆ’1)0+4(0)βˆ’2=ο€Ή0+2+3ο€Ή0+2(0βˆ’1)+(𝐢(0)+𝐷)(0βˆ’1)βˆ’2=2+3(2)(βˆ’1)+(𝐷)(βˆ’1)=2βˆ’6+𝐷∴𝐷=2.

We can now substitute any other value for π‘₯ to determine 𝐢. Say π‘₯=2, which gives π‘₯+4π‘₯βˆ’2=ο€Ήπ‘₯+2+3ο€Ήπ‘₯+2(π‘₯βˆ’1)+(𝐢π‘₯+2)(π‘₯βˆ’1)2+4(2)βˆ’2=ο€Ή2+2+3ο€Ή2+2(2βˆ’1)+((2)𝐢+2)(2βˆ’1)22=2𝐢+26∴𝐢=βˆ’2.

We have found that π‘₯+4π‘₯βˆ’2(π‘₯βˆ’1)(π‘₯+2)=1(π‘₯βˆ’1)+3(π‘₯βˆ’1)+2βˆ’2π‘₯(π‘₯+2).

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