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Partial Fractions

Partial Fractions

The process of partial fractions is a method of decomposing rational functions of the form $\frac{P(x)}{Q(x)}$. In the simplest form, we decompose functions into the form $$ \frac{P(x)}{Q(x)} \equiv \frac{A_1}{x-\lambda_1} + \frac{A_2}{x-\lambda_2} + \frac{A_1}{x-\lambda_3} + \frac{A_1}{x-\lambda_4} +\dots $$ Such a decomposition requires that all the roots of $Q(x)$ are real. To achieve this decomposition, we cross multiply by $Q(x)$ $$ P(x) \equiv Q(x)\left(\frac{A_1}{x-\lambda_1} + \frac{A_2}{x-\lambda_2} + \frac{A_1}{x-\lambda_3} + \frac{A_1}{x-\lambda_4} +\dots\right) $$ since this is true for all values of $x$, it is true for the values $\lambda_1,\lambda_2,\dots$. If we substitute in these values, we arrive at simple equations we can solve to determine the values of $A_i$

Partial Fractions 1

Decompose the following rational function using the partial fractions method $$ \frac{2x + 4}{x^2 - 13x + 30} $$
solution - press button to display

Our first step is to decompose the denominator. $x^2 - 13x + 30 = (x-3)(x-10)$ $$ \begin{array}{rcl} \frac{2x + 4}{x^2 - 13x + 30} &=& \frac{A}{x-3} + \frac{B}{x-10}\\ 2x+4 &= &A(x-10) + B(x-3) \end{array} $$ Taking $x = 10$, gives $24 = 7B$ giving $B = \frac{24}{7}$ Taking $x = 3$, gives $10 = -7A$ giving $A = -\frac{10}{7}$ Therefore $$ \frac{2x + 4}{x^2 - 13x + 30} = \frac{-10}{7(x-3)} + \frac{24}{7(x-10)}\\ $$

Partial Fractions 2

Using the by parts method, decompose the following rational function $$ f(x) = \frac{13x^2 +8x -72}{x^3 + x^2 -14x -24 } $$
solution - press button to display

In the absence of any other information, I would suggest testing the factors of $\pm 24$ as possible roots of $x^3 + x^2 - 14x -24 =0$. As it happens, x = -2 is one such solution.

From this it is now easy to completely factorise the denominator as $x^3 + x^2 - 14x - 24 = (x+2)(x+3)(x-4)$.

 

The function $f(x)$ can be initially rewritten as $$ f(x) = \frac{13x^2 +8x -72}{(x+2)(x+3)(x-4)} \equiv \frac{A}{x+2} + \frac{B}{x+3} + \frac{C}{x-4} $$

Multiplying both sides of the equivalence gives us the new equivalence $$13x^2 + 8x - 72 \equiv A(x+3)(x-4) + B(x+2)(x-4) + C(x+2)(x+3)$$ We can now choose any values of $x$ to help determine the parameters $A,B,C$.

Let $x = 4$, then we get $$ \begin{array}{rcl}13(4)^2 + 8(4) - 72 &=& 168\\  168 &=& A(7)(0) + B(2)(0) + 42C \\ C&=&4\end{array} $$

Let $x=-3$ then we get $$ \begin{array}{rcl}13(-3)^2 + 8(-3) - 72 &=& 21\\  21 &=& A(0)(-7) + B(-1)(-7) + C(-1)(0) \\ B&=&3\end{array} $$

Let $x=-2$ then we get $$ \begin{array}{rcl}13(-2)^2 + 8(-2) - 72 &=& -36\\  -36 &=& A(1)(-6) + B(0)(-6) + C(0)(1) \\ A&=&6\end{array} $$

The function $f(x)$ can therefore be rewritten as $$f(x) = \frac{6}{x+2} + \frac{3}{x+3} + \frac{4}{x-4}  $$

Partial Fractions 3

Decompose the function $g(x) = \frac{20}{(x+1)(x^2 + 4)}$ using the method of partial fractions.

solution - press button to display

As $x^2+4$ is irreducible over $\mathbb{R}$, then our decomposition will be of the form $$ \frac{20}{(x+1)(x^2+4)} \equiv \frac{A}{x+1} + \frac{Bx+C}{x^2+4} $$

Cross multiplication yields $$ 20 = A(x^2+4) + (Bx+C)(x+1) $$

The standard method can once again be applied, let us take $x=-1$, $x=0$, $x=1$ as our values for $x$, then we get $$x = -1 \Rightarrow 20=5A \Rightarrow A = 4$$ $$x=0 \Rightarrow 20 = 4A + C \Rightarrow C = 4$$ $$x = 1 \Rightarrow 20 = 5A + 2(B+C) \Rightarrow B = -4$$

So the function $g(x)$ can be rewritten as $$ g(x)  \equiv \frac{4}{x+1} + \frac{-4x+4}{x^2+4} $$