Partial fractions of $\frac{-5x+19}{(x-1/2)(x+1/3)}$

Question

Alright, I need to find the partial fractions for the expression above. I have tried writing this as $$\frac{a}{x-1/2}+\frac{b}{x+1/3}$$ but the results give me $a=25.8$ and $b=-20.8$, which are slightly wrong because they give me $5x+19$ instead of $-5x+19$. Can you please help? Thanks a lot

Answer 1

For a specific type of fractions, there is an easier way - it's my favourite - to find partial fractions. If a given fraction satisfies the following 2 conditions, then it becomes amazingly easy to separate it into partial fractions:

• The degree of numerator is $<$ the degree of denominator
• The denominator can be expressed as a product of linear and non-repeating factors (that is, each factor in the denominator is raised to the power 1)

Your question satisfies these two conditions. So, here's the method: (I'll explain using an example)

Consider $ \dfrac { x - 2 } {(x-7)(x + 8)} $. Now, notice that the denominator vanishes at $ x = 7 $ and $ x = -8 $. So, put $x = 7$ everywhere but there where the denominator's factor becomes zero.

That is, write this: $$ \dfrac {7-2} { (x - 7)(7 + 8) } = \dfrac 5 {15(x-7)} $$

Now substitute $ x = -8$ everywhere but there where the denominator's factor becomes zero.

That is, write this: $$ \dfrac { -8 - 2 } {(-8-7)(x - 8)} = \dfrac {10} {15(x+8)} $$

So, your answer is the sum of these two terms. That is:

$$ \dfrac { x - 2 } {(x-7)(x + 8)} = \dfrac 5 {15(x-7)} + \dfrac {10} {15(x+8)} $$

You can check. It works.

So your fraction can be written as:

$$ \dfrac{-5x+19}{(x-1/2)(x+1/3)} = \dfrac { 16.5 } { 5/6(x - 1/2) } - \dfrac { 21.67 } { 1/5(x - 1/3) } $$

Answer 2

Either you've started from the wrong equations, or you've made a calculation mistake (don't worry, happens to all of us sometimes). When you write $$\frac{a}{x-\frac{1}{2}}+\frac{b}{x+\frac{1}{3}}=\frac{-5x+19}{(x-\frac{1}{2})(x+\frac{1}{3})}$$ you get $$\frac{a(x+\frac{1}{3})}{(x-\frac{1}{2})(x+\frac{1}{3})}+\frac{b(x-\frac{1}{2})}{(x-\frac{1}{2})(x+\frac{1}{3})}=\frac{ax+bx+\frac{a}{3}-\frac{b}{2}}{(x-\frac{1}{2})(x+\frac{1}{3})}=\frac{-5x+19}{(x-\frac{1}{2})(x+\frac{1}{3})}$$ so that the correct equations to solve are $$\begin{align*} a+b&=-5\\[0.1in] \tfrac{a}{3}-\tfrac{b}{2}&=19 \end{align*}$$