Partial Fractions Done Right
The Case of Simple Real Roots
The partial partial fractions form for a simple real root $a$ is:
\[
\frac{p(x)}{(x-a)\phi(x)} = \frac{A}{x-a} + \frac{q(x)}{\phi(x)}.
\]
Multiplying through by $x-a$
\[
\frac{p(x)}{\phi(x)} = A + \frac{(x-a)q(x)}{\phi(x)}
\]
then substituting $x=a$ we get
\[
A = \frac{p(a)}{\phi(a)}.
\]
A good way to view this result is to imagine that we "hide" the
$(x-a)$ in the original fraction's denominator and evaluate what remains
at $x=a$:
\[
A =
\frac{p(x)}{\bbox[#ffdddd]{\color{#999999}{(x-a)}}\phi(x)} \bigg|_{x=a}
=
\frac{p(a)}{\phi(a)}.
\]
Compute $A$ in the partial partial fractions expansion
\[
\frac{4x^2 + 2x + 1}{(x-1)(x-2)^2(x^2+x+1)}
= \frac{A}{x-1} + \frac{q(x)}{(x-2)^2(x^2+x+1)} .
\]
We have:
\[
A = \frac{4x^2 + 2x + 1}{\bbox[#ffdddd]{\color{#999999}{(x-1)}}(x-2)^2(x^2+x+1)} \bigg|_{x=1} = \frac{7}{3},
\]
therefore
\[
\frac{4x^2 + 2x + 1}{(x-1)(x-2)^2(x^2+x+1)}
= \frac{7}{3(x-1)} + \frac{q(x)}{(x-2)^2(x^2+x+1)} .
\]
Compute the
full partial fractions expansion
\[
\frac{2x^2+x+1}{(x-1)(x-2)(x-3)}
= \frac{A_1}{x-1}
+ \frac{A_2}{x-2}
+ \frac{A_3}{x-3}.
\]
We have:
\begin{align*}
A_1 &= \frac{2x^2+x+1}{\bbox[#ffdddd]{\color{#999999}{(x-1)}}(x-2)(x-3)} \bigg|_{x=1} = 2, \\
A_2 &= \frac{2x^2+x+1}{(x-1)\bbox[#ffdddd]{\color{#999999}{(x-2)}}(x-3)} \bigg|_{x=2} = -11, \\
A_3 &= \frac{2x^2+x+1}{(x-1)(x-2)\bbox[#ffdddd]{\color{#999999}{(x-3)}}} \bigg|_{x=3} = 11,
\end{align*}
therefore:
\[
\frac{2x^2+x+1}{(x-1)(x-2)(x-3)}
= \frac{2}{x-1}
- \frac{11}{x-2}
+ \frac{11}{x-3}.
\]
Test Yourself
-
Show that:
\[
\frac{3x^2+2x+1}{(x-1)(x-2)^2(1+x+x^2)}
=
\frac{2}{x-1} + \frac{q(x)}{(x-2)^2(1+x+x^2)}.
\]
-
Show that:
\[
\frac{1}{x^2-1} = \frac{1}{2(x-1)} - \frac{1}{2(x+1)}.
\]
-
Show that:
\[
\frac{x+1}{x^2+5x+6} = -\frac{1}{x+2} + \frac{2}{x+3}.
\]
-
Assume $a$, $b$, $c$ are distinct. Show that:
\[
\frac{1}{(x-a)(x-b)(x-c)}
=
\frac{1}{(b-a)(c-a)(x-a)}
+
\frac{1}{(c-b)(a-b)(x-b)}
+
\frac{1}{(a-c)(b-c)(x-c)}.
\]
-
Assume $a \ne b$. Show that:
\[
\frac{r x + s}{(x+a)(x+b)}
=
\frac{\frac{s-ra}{b-a}}{x+a}
+
\frac{\frac{s-rb}{a-b}}{x+b}.
\]
Author: Rouben Rostamian