4.8 Partial fractions

When we add fractions (here a,b,c,da,b,c,d\in\mathbb{R}77 7 Even though they’re in the set of real numbers, they don’t have to be written down directly as numbers (like 1.21.2 or 2322^{3}2) - they could be algebraic expressions (such as x3x-3), so long as when the value of those algebraic expressions is computed, the results are real numbers.), something like this happens

ab+cd\displaystyle\frac{a}{b}+\frac{c}{d} =adbd+cbdb\displaystyle=\frac{ad}{bd}+\frac{cb}{db} (4.70)
=ad+cbdc\displaystyle=\frac{ad+cb}{dc} (4.71)

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


We can break it apart. We can use Equation 4.71 to predict (if we set b=x+4b=x+4 and d=x8d=x-8, and ab+cb=x+3ab+cb=x+3) that "splitting up" (aka partial fraction) will look like


To solve this, we take advantage of the fact that because the values of aa and bb are constants88 8 i.e. their value is always the same. This means that they must be the same for every value of xx, so if we plug in specific values of xx, and then find what aa and bb are given those values of xx we know that those values of aa and bb will be true for all values of xx.

times by (x+4)(x8), then x+3=a(x8)+b(x+4)\displaystyle\text{times by $(x+4)(x-8)$, then }x+3=a(x-8)+b(x+4)
Find find aa
set x=4, then 4+3=a(48)+b(4+4)\displaystyle\text{set $x=-4$, then }-4+3=a(-4-8)+b(-4+4)
a=112\displaystyle\implies a=\frac{1}{12}
Then find bb
set x=8, then 8+3=a(88)+b(8+4)\displaystyle\text{set $x=8$, then }8+3=a\cdot(8-8)+b(8+4)
12b=11\displaystyle\implies 12b=11
b=1112\displaystyle\implies b=\frac{11}{12}
Thus overall
x+3(x+4)(x8)=112x+4+1112x8\displaystyle\frac{x+3}{(x+4)(x-8)}=\frac{\frac{1}{12}}{x+4}+\frac{\frac{11}{1% 2}}{x-8}

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

112(x+4)+1112(x8)\displaystyle\frac{1}{12(x+4)}+\frac{11}{12(x-8)} =(x8)+11(x+4)12(x+4)(x8)\displaystyle=\frac{(x-8)+11(x+4)}{12(x+4)(x-8)}
Which is what we started out with!

There’s another case, however, which we haven’t looked at yet; what if there’s a repeated root 99 9 Remember, from the factor theorem (see above if not) that roots, i.e. when a polynomial is zero, are also factors (e.g. x+4x+4 is a factor of (x+4)(x8)(x+4)(x-8), and 44 is a root.).

In this case we have something like


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

x+3=A+B\displaystyle x+3=A+B

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

x+3(x+4)2\displaystyle\frac{x+3}{(x+4)^{2}} =(x+4)1(x+4)2\displaystyle=\frac{(x+4)-1}{(x+4)^{2}} (4.72)

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


We could try something like


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

A(x+4)2+Bx8\displaystyle\frac{A}{(x+4)^{2}}+\frac{B}{x-8} =A(x8)+B(x+4)2(x+4)2(x+3)\displaystyle=\frac{A(x-8)+B(x+4)^{2}}{(x+4)^{2}(x+3)}

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


We can "compare coefficients"1010 10 Note: this is explored further in the (not yet written) ”algebra techniques” section. and obtain that

{Bx2=0x2(A+8B)x=x(16B8A)=3\begin{cases}Bx^{2}=0x^{2}\\ (A+8B)x=x\\ (16B-8A)=3\end{cases}

Which simplifies to

{B=0(A+8B)=1(16B8A)=3\begin{cases}B=0\\ (A+8B)=1\\ (16B-8A)=3\end{cases}

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

{A=18A=3A=38\begin{cases}A=1\\ -8A=3\implies A=-\frac{3}{8}\end{cases}

Which means that the equations are inconsistent 1111 11 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+BAx+B.

x+3=(Ax+B)(x8)+C(x+4)2\displaystyle\implies x+3=(Ax+B)(x-8)+C(x+4)^{2}
Find C first because it’s the easiest one to find.\displaystyle\textbf{Find $C$ first because it's the easiest one to find}.
If we set x=8x=8, then (x8)(x-8), and thus the whole left-hand part will be zero
11=(8A+B)(88)+122C\displaystyle 11=(8A+B)(8-8)+12^{2}C
C=11144\displaystyle C=\frac{11}{144}
Now find AA and BB
x+3=(Ax+B)(x8)+11144(x+4)2\displaystyle x+3=(Ax+B)(x-8)+\frac{11}{144}(x+4)^{2}
x+3=Ax28Ax+Bx8B+11144x2+88144x+176144\displaystyle x+3=Ax^{2}-8Ax+Bx-8B+\frac{11}{144}x^{2}+\frac{88}{144}x+\frac{1% 76}{144}
(11144+A)x2+(B1+88144)x+(1761448B3)=0\displaystyle(\frac{11}{144}+A)x^{2}+(B-1+\frac{88}{144})x+(\frac{176}{144}-8B% -3)=0
A+11144=0A=11144\displaystyle A+\frac{11}{144}=0\implies A=-\frac{11}{144}
B1+88144=0B=188144B=56144\displaystyle B-1+\frac{88}{144}=0\implies B=1-\frac{88}{144}\implies B=\frac{% 56}{144}

There’s a caveat to all this, unfortunately! If the degree 1212 12 i.e. the highest power of the polynomial - e.g. for x2+x+1x^{2}+x+1 the degree would be 22 and for z3+8z+5z^{3}+8z+5 it would be 33 of the numerator is equal to or greater than the numerator, we first have to divide (using polynomial long division1313 13 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).