Question about partial fractions with repeated linear factors

Question

I am teaching A Level Maths, and have a question about partial fractions.

When there is a denominator with a repeated linear factor, such as the following example:

$$\frac{2x+9}{(x-5)(x+3)^2}$$

the exercise book, and all similar examples, say that you should split it into 3 partial fractions of:

$$\frac{A}{(x-5)}+\frac{B}{(x+3)}+\frac{C}{(x+3)^2}$$

I understand why you can't just set it up as:

$$\frac{A}{(x-5)}+\frac{B}{(x+3)}+\frac{C}{(x+3)}$$

But I don't understand why you couldn't just break it into 2 partial fractions of:

$$\frac{A}{(x-5)}+\frac{B}{(x+3)^2}$$

Is anyone able to explain why all the questions like this insist on breaking it into 3 fractions, rather than 2? Especially since there is no avoiding a fraction with a quadratic denominator, no matter which way you do it.

In your final expression, $\frac{A}{(x-5)}+\frac{Bx+D}{(x+3)^2}$ works — Will Jagy, May 03 '23 at 16:29
Does this answer your question? Repeated linear factors in partial fractions - found using a site search. There are also other duplicate, or at least closely related questions, e.g., Partial Fraction Decomp. Why repeated factors. And why the (Cx+D) for quadratics?, Derivation of the general forms of partial fractions, ... — John Omielan, May 03 '23 at 16:33
Also see https://math.stackexchange.com/questions/3172683/why-partial-fraction-decomposition-of-frac1s2s2-is-fracas-frac?noredirect=1&lq=1 — A. Goodier, May 03 '23 at 16:36
(cont.) The numerators in partial fraction decomposition involving repeated factors, Why do repeated linear factors have to be dealt with in this way?, etc. — John Omielan, May 03 '23 at 16:37

score 0 · Answer 1 · answered May 03 '23 at 16:40

If the fraction is written as the ratio of two polynomials

$$\frac{P(x)}{Q(x)}$$

where the highest power of $P(x)$ is less than the highest power of $Q(x)$. Suppose you factorize the $Q(x)$ into small factors, for example, in your case, $Q(x)=(x-5)(x+3)^2$, so in general, you get

$$\frac{P(x)}{Q(x)}=\frac{A(x)}{x-5}+\frac{B(x)}{(x+3)^2}$$

Note that the highest power of the numerator need to be one-order less than the highest power of the denominator, therefore,

$A(x)=C_1=constant$

$B(x)=C_2x+C_3$

Next, you can write $B(x)=C_2(x+3)+C_4$, where $C_4=C_3-3C_2$, hence, we get

$$\frac{P(x)}{Q(x)}=\frac{A(x)}{x-5}+\frac{B(x)}{(x+3)^2}=\frac{C_1}{x-5}+\frac{C_2(x+3)+C_4}{(x+3)^2}=\frac{C_1}{x-5}+\frac{C_2}{(x+3)}+\frac{C_4}{(x+3)^2}$$

Question about partial fractions with repeated linear factors

1 Answers1