A calculus of the absurd

4.8 Partial fractions

When we add fractions (here $$a, b, c, d \in \mathbb {R}$$1616 Even though they’re in the set of real numbers, they don’t have to be written down directly as numbers (like $$1.2$$ or $$2^32$$) - they could be algebraic expressions (such as $$x-3$$), so long as when the value of those algebraic expressions is computed, the results are real numbers.), something like this happens

\begin{align} \frac {a}{b} + \frac {c}{d} & = \frac {ad}{bd} + \frac {cb}{db} \\ & = \frac {ad + cb}{dc} \label {generic fraction addition} \end{align}

Sometimes we have fractions which are the "added together" form, which we’d like to turn into the "not added together" form. For example in the case of this fraction

\begin{equation*} \frac {x+3}{(x+4)(x-8)} \end{equation*}

We can break it apart. We can use Equation 4.71 to predict (if we set $$b=x+4$$ and $$d=x-8$$, and $$ab + cb = x+3$$) that "splitting up" (aka partial fraction) will look like

\begin{equation*} \frac {x+3}{(x+4)(x-8)} = \frac {a}{x+4} + \frac {b}{x-8} \end{equation*}

To solve this, we take advantage of the fact that because the values of $$a$$ and $$b$$ are constants1717 i.e. their value is always the same. This means that they must be true for every value of $$x$$, so if we plug in specific values of $$x$$, and then find what $$a$$ and $$b$$ are given those values of $$x$$ we know that those values of $$a$$ and $$b$$ will be true for all values of $$x$$.

\begin{align*} & \frac {x+3}{(x+4)(x-8)} = \frac {a}{x+4} + \frac {b}{x-8} \\ & \text {times by $(x+4)(x-8)$, then }x+3 = a (x-8) + b (x+4) \\ & \textbf {Find find $a$} \\ & \text {set $x=-4$, then } -4 + 3 = a(-4 -8) + b(-4 + 4) \\ & -1 = -12 a \\ & a = \frac {1}{12} \\ & \textbf {Then find $b$} \\ & \text {set $x=8$, then } 8 + 3 = a \cdot (8 -8) + b (8 + 4) \\ & 12 b = 11 \\ & b = \frac {11}{12} \\ & \textbf {Thus overall} \\ & \frac {x+3}{(x+4)(x-8)} = \frac {\frac {1}{12}}{x+4} + \frac {\frac {11}{12}}{x-8} \\ & \frac {x+3}{(x+4)(x-8)} = \frac {1}{12(x+4) } + \frac {11}{12(x-8)} \end{align*}

We can also check the answer by adding the fractions back together again:

\begin{align*} \frac {1}{12(x+4) } + \frac {11}{12(x-8)} & = \frac {(x-8) + 11(x+4)}{12(x+4)(x-8)} \\ & = \frac {12x - 36}{12(x+4)(x-8)} \\ & = \frac {x - 3}{(x+4)(x-8)} \\ & \quad \text {Which is what we started out with!} \end{align*}

There’s another case, however, which we haven’t looked at yet; what if there’s a repeated root 1818 Remember, from the factor theorem (see above if not) that roots, i.e. when a polynomial is zero, are also factors (e.g. $$x+4$$ is a factor of $$(x+4)(x-8)$$, and $$4$$ is a root.).

In this case we have something like

\begin{equation*} \frac {x+3}{(x+4)^2} \end{equation*}

if we try to apply the previous technique, we do something like:

\begin{align*} & \frac {x+3}{(x+4)^2} = \frac {A}{(x+4)^2} + \frac {B}{(x+4)^2} \\ & x + 3 = A + B \end{align*}

This is impossible to solve (because clearly values of $$A$$ and $$B$$ which satisfy one value of $$x$$ are not going to satisfy all values of $$x$$) - oops! The trick here is to do some clever algebra. We can rewrite $$x+3$$ in terms of $$x+4$$ (because $$x+3 = (x+4) - 1$$), and from there it becomes much easier.

\begin{align} \label {partial fractions with repeated root with specific values} \frac {x+3}{(x+4)^2} & = \frac {(x+4)-1}{(x+4)^2} \\ & = \frac {x+4}{(x+4)^2} + \frac {-1}{(x+4)^2} \notag \\ & = \frac {1}{x+4} - \frac {1}{(x+4)^2} \notag \end{align}

The final thing is what happens when some other terms are mixed in, as in

\begin{equation*} \frac {x+3}{(x+4)^2(x-8)} \end{equation*}

We could try something like

\begin{equation*} \frac {x+3}{(x+4)^2(x-8)} = \frac {A}{(x+4)^2} + \frac {B}{x-8} \end{equation*}

however, this fails pretty quickly if we try to add the right-hand side back together

\begin{align*} \frac {A}{(x+4)^2} + \frac {B}{x-8} & = \frac {A(x-8) + B(x+4)^2}{(x+4)^2(x+3)} \\ & = \frac {Ax - 8A + Bx^2 + 8Bx + 16 B}{(x+4)^2(x+3)} \\ & = \frac {Ax - 8A + Bx^2 + 8Bx + 16 B}{(x+4)^2(x+3)} \\ & = \frac {Bx^2 + (A + 8B)x + (16B - 8A)}{(x+4)^2(x+3)} \end{align*}

We know that when we add the fractions together, we should get $$x+3$$ as the numerator. Therefore the numerator of the partial fractions added together should be the same as the numerator of the original fraction.

\begin{equation*} Bx^2 + (A + 8B)x + (16B - 8A) = x + 3 \end{equation*}

We can "compare coefficients"1919 Note: this is explored further in the (not yet written) "algebra techniques" section. and obtain that

\begin{equation*} \begin{cases} Bx^2 = 0x^2 \\ (A + 8B)x = x \\ (16B - 8A) = 3 \end {cases} \end{equation*}

Which simplifies to

\begin{equation*} \begin{cases} B = 0 \\ (A + 8B) = 1 \\ (16B - 8A) = 3 \end {cases} \end{equation*}

If we substitute $$B=0$$ into the other equations, we get that

\begin{equation*} \begin{cases} A = 1 \\ - 8A = 3 \implies A = -\frac {3}{8} \end {cases} \end{equation*}

Which means that the equations are inconsistent 2020 i.e. are no solutions to the equations - more on inconsistent/consistent equations in the "linear algebra" section (though that part of the "linear algebra" section is yet to be written), and that there are no solutions satisfying all the equations at the same time.

Instead, we need to try to express this in a form similar to Equation 4.72, i.e. something like $$Ax + B$$.

\begin{align*} & \frac {x+3}{(x+4)^2(x-8)} = \frac {Ax + B}{(x+4)^2} + \frac {C}{x-8} \\ & x + 3 = (Ax + B)(x-8) + C(x+4)^2 \\ & \textbf {Find $C$ first because it's the easiest one to find} \\ & 11 = (8A + B)(8 - 8) + 12^2 C \\ & C = \frac {11}{144} \\ & \textbf {Now find $A$ and $B$} \\ & x + 3 = (Ax + B)(x-8) + \frac {11}{144} (x+4)^2 \\ & x + 3 = Ax^2 -8 Ax + Bx - 8B + \frac {11}{144} x^2 + \frac {88}{144} x + \frac {176}{144} \\ & (\frac {11}{144} + A)x^2 + (B - 1 + \frac {88}{144})x + (\frac {176}{144}-8B-3) = 0 \\ & A + \frac {11}{144} = 0 \implies A = - \frac {11}{144} \\ & B - 1 + \frac {88}{144} = 0 \implies B = 1 - \frac {88}{144} \implies B = \frac {56}{144} \end{align*}

There’s a caveat to all this, unfortunately! If the degree 2121 i.e. the highest power of the polynomial - e.g. for $$x^2 + x + 1$$ the degree would be $$2$$ and for $$z^3 + 8z + 5$$ it would be $$3$$ of the numerator is equal to or greater than the numerator, we first have to divide (using polynomial long division2222 Also in the yet-to-be-written algebra section) the numerator by the denominator to get an expression which isn’t a fraction, plus a remainder (where the degree of the numerator is less than the denominator).