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Can we use either $$ \frac{x-5}{(x-2)^2 (x+1)} = \frac{A}{(x-2)^2} + \frac{B}{(x-2)} + \frac{C}{(x+1)} $$

or equivalently,

$$ \frac{x-5}{(x-2)^2 (x+1)} = \frac{A'x+B'}{(x-2)^2} + \frac{C}{(x+1)} $$

since $(x-2)^2$ is a quadratic factor. My teacher only told us to "use the equation inside the bracket", what is the reason behind this?

cdeamaze
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I'm a lightbulb
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3 Answers3

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The question as stated was incorrect. The correct form with 3 unknowns A, B and C

$$f(x) = (x−5)/[(x−2)^2(x+1)]= A/(x−2)^2 + B/(x−2) + C/(x+1).\tag {a}$$

which is equivalent to

$$f(x) = (Bx+(A-2B))/(x−2)^2 + C/(x+1).\tag {b}$$

where we have A' = B and B' = A-2B. are defined as an alternative form eq (b)

Solution for A, B, and C are obtained from eqs (2), (3), and (1).

$$C = {(x-5)/[(x-2)^2]}|_{x=-1} = - 6/9 = -2/3.\tag {1}$$

$$A = {(x-5)/(x+1)}|_{x=2} = - 1.\tag {2}$$

(x−5)/[(x−2)^2(x+1)] + 1/(x-2)^2 = [(x-5)+(x+1)]/[(x−2)^2(x+1)] = 2/[(x-2)(x+1)]

Hence we have 2/[(x-2)(x+1)] = B/(x-2) + C/(x+1)

Solve for B,

$$B = [2/(x+1)]|_{x=2} = 2/3.\tag {3}$$

Substituting the known coefficients A, B and C, eq (a) now becomes

$$f(x) = (x−5)/[(x−2)^2(x+1)] = - 1/(x−2)^2 + 2/[3(x−2)] - 2/[3(x+1)].\tag {a}$$

cdeamaze
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$$\frac{x-5}{(x-2)^2(x+1)}=\frac{Ax+B}{(x-2)^2}+\frac{C}{x+1}+\frac{D}{x-2}\\\to(Ax+B)(x+1)+C(x-2)^2+D(x-2)(x+1)=x-5$$

$$x=2\implies3(2A+B)=-3\implies2A+B=-1\tag I$$ $$x=-1\implies9C=-6\implies C=-\frac23\tag{ II}$$ $$x=0\implies B-\frac83-2D=-5\implies B=2D-\frac73\implies A+D=\frac23\tag{III}$$ It is impossible from here to exactly determine $A,B,D$ as the solution is not unique (indeed, any choice is valid so long as $B$ and $D$ meet $(\text I)$ and $(\text{III})$). So, we take the simplest decomposition, where $A=0$.

Rhys Hughes
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  • Not true here. $$\frac{Ax-(1+2A)}{(x-2)^2}-\frac{2}{3(x-1)}+\frac{2-3A}{3(x-2)}$$ works for any choice of $A$. You can check that here. Indeed, the problem here is that inserting $B$ and $D$ in terms of $A$ in the first polynomial zeroes it out regardless of the choice of $A$ – Rhys Hughes May 25 '21 at 13:35
  • There is a unique solution for A, B and C, where A, B and C are defined in eq(a) of my post. Please see my post for details. Every partial fraction expansion has a unique solution whether eq(a) or eq(b) in my post is used. – cdeamaze May 27 '21 at 07:11
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Here is an analogy that might help.

When you write $\dfrac{78}{10^2} = \dfrac{78}{100}$ as a decimal fraction, you get $0.78$, which equals $\dfrac{7}{10} + \dfrac{8}{100} = \dfrac{7}{10} + \dfrac{8}{10^2}$, where the numerators are both non negative integers less than $10$.

In very much the same sort of reasoning, when you $``\text{expand}"$, $\dfrac {P(x)}{(x-2)^2}$ you will get $\dfrac{A}{x-2} + \dfrac{B}{(x-2)^2}$ where the numerators both have a degree less than $1$, (that is to say, they are constants), which is the degree of $x-2$

So, yes, $\dfrac{x-5}{(x-2)^2 (x+1)} = \dfrac{Ax+B}{(x-2)^2} + \dfrac{C}{(x+1)}$ is correct.

Steven Alexis Gregory
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