# 4.8 Partial fractions

When we add fractions (here $a,b,c,d\in\mathbb{R}$^{7}^{7}
7
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 $2^{3}2$) - 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

$\displaystyle\frac{a}{b}+\frac{c}{d}$ | $\displaystyle=\frac{ad}{bd}+\frac{cb}{db}$ | (4.70) | ||

$\displaystyle=\frac{ad+cb}{dc}$ | (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 $b=x+4$ and $d=x-8$, and $ab+cb=x+3$) that "splitting up" (aka partial fraction) will look like

To solve this, we take advantage of the fact that because the values of $a$ and
$b$ are constants^{8}^{8}
8
i.e. their value is always the same. This means
that they must be the same 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 $x$.

$\displaystyle\frac{x+3}{(x+4)(x-8)}=\frac{a}{x+4}+\frac{b}{x-8}$ | ||

$\displaystyle\text{times by $(x+4)(x-8)$, then }x+3=a(x-8)+b(x+4)$ | ||

Find find $a$ | ||

$\displaystyle\text{set $x=-4$, then }-4+3=a(-4-8)+b(-4+4)$ | ||

$\displaystyle\implies-1=-12a$ | ||

$\displaystyle\implies a=\frac{1}{12}$ | ||

Then find $b$ | ||

$\displaystyle\text{set $x=8$, then }8+3=a\cdot(8-8)+b(8+4)$ | ||

$\displaystyle\implies 12b=11$ | ||

$\displaystyle\implies b=\frac{11}{12}$ | ||

Thus overall | ||

$\displaystyle\frac{x+3}{(x+4)(x-8)}=\frac{\frac{1}{12}}{x+4}+\frac{\frac{11}{1% 2}}{x-8}$ | ||

$\displaystyle\frac{x+3}{(x+4)(x-8)}=\frac{1}{12(x+4)}+\frac{11}{12(x-8)}$ |

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

$\displaystyle\frac{1}{12(x+4)}+\frac{11}{12(x-8)}$ | $\displaystyle=\frac{(x-8)+11(x+4)}{12(x+4)(x-8)}$ | ||

$\displaystyle=\frac{12x-36}{12(x+4)(x-8)}$ | |||

$\displaystyle=\frac{x-3}{(x+4)(x-8)}$ | |||

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 ^{9}^{9}
9
Remember, from the factor theorem (see above if not)
that roots, i.e. when a polynomial is zero, are also factors (e.g. $x+4$ is a
factor of $(x+4)(x-8)$, and $4$ is a root.).

In this case we have something like

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

$\displaystyle\frac{x+3}{(x+4)^{2}}=\frac{A}{(x+4)^{2}}+\frac{B}{(x+4)^{2}}$ | ||

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

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.

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

$\displaystyle=\frac{x+4}{(x+4)^{2}}+\frac{-1}{(x+4)^{2}}$ | ||||

$\displaystyle=\frac{1}{x+4}-\frac{1}{(x+4)^{2}}$ |

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

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

$\displaystyle=\frac{Ax-8A+Bx^{2}+8Bx+16B}{(x+4)^{2}(x+3)}$ | |||

$\displaystyle=\frac{Ax-8A+Bx^{2}+8Bx+16B}{(x+4)^{2}(x+3)}$ | |||

$\displaystyle=\frac{Bx^{2}+(A+8B)x+(16B-8A)}{(x+4)^{2}(x+3)}$ |

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.

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

Which simplifies to

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

Which means that the equations are inconsistent ^{11}^{11}
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 $Ax+B$.

$\displaystyle\frac{x+3}{(x+4)^{2}(x-8)}=\frac{Ax+B}{(x+4)^{2}}+\frac{C}{x-8}$ | ||

$\displaystyle\implies x+3=(Ax+B)(x-8)+C(x+4)^{2}$ | ||

$\displaystyle\textbf{Find $C$ first because it's the easiest one to find}.$ | ||

If we set $x=8$, then $(x-8)$, and thus the whole left-hand part will be zero | ||

$\displaystyle 11=(8A+B)(8-8)+12^{2}C$ | ||

$\displaystyle C=\frac{11}{144}$ | ||

Now find $A$ and $B$ | ||

$\displaystyle x+3=(Ax+B)(x-8)+\frac{11}{144}(x+4)^{2}$ | ||

$\displaystyle x+3=Ax^{2}-8Ax+Bx-8B+\frac{11}{144}x^{2}+\frac{88}{144}x+\frac{1% 76}{144}$ | ||

$\displaystyle(\frac{11}{144}+A)x^{2}+(B-1+\frac{88}{144})x+(\frac{176}{144}-8B% -3)=0$ | ||

$\displaystyle A+\frac{11}{144}=0\implies A=-\frac{11}{144}$ | ||

$\displaystyle B-1+\frac{88}{144}=0\implies B=1-\frac{88}{144}\implies B=\frac{% 56}{144}$ |

There’s a caveat to all this, unfortunately! If the degree
^{12}^{12}
12
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 division^{13}^{13}
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).