# Repeated Linear Factors in denominator of fraction e.g. $\frac{2x^2 + 2x + 18}{x(x-3)^2}$

I have the following fraction below and I must find the partial fraction decomposition. $$\frac{2x^2 + 2x + 18}{x(x-3)^2}$$

Now, I thought I could simplify this into the following...

$$\frac{2x^2 + 2x + 18}{x(x-3)^2} = \frac{A}{x} + \frac{B}{(x-3)^2}$$

However, my professor said that this is incorrect, and instead it should be written as:

$$\frac{2x^2 + 2x + 18}{x(x-3)^2} = \frac{A}{x} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$$

I'm confused.

I can't seem to find an explanation for why my method is wrong and why my professor's solution is correct, using algebraic reasoning.

Is there a reason why this is the case?

• Perhaps it's clearer to combine the $(x-3)$ terms as $\frac {bx+c}{(x-3)^2}$. Note: my $b,c$ aren't quite the same as your professor's $B,C$. There's a small shift. In general, the numerator should have degree one less than the denominator. – lulu Aug 30 '17 at 20:58
• To understand this, suppose the original expression was $\frac {x+1}{(x+3)^2}$. Would you try to write this as $\frac A{(x+3)^2}$ ? – lulu Aug 30 '17 at 20:59
• @lulu I will add my own answer shortly now that I have understood your comment! Many thanks! – vik1245 Aug 30 '17 at 21:05
• If you write $\frac{B}{x-3}$ as $\frac{B(x-3)}{(x-3)^2}$, and combine with the third fraction, it becomes clear what Lulu meant. – imranfat Aug 30 '17 at 21:12

If we accept your solution and try to find the coefficients $A$ and $B$ by identification after we have reduced the r.h.s. to the same denominator, wu obtain $$A(x-3)^2+Bx=2x^2+2x+18\iff Ax^2+(B-6A)x+9A=2x^2+2x+18,$$ whence the linear system \begin{cases}A=2,\\B-6A=2,\\9A=18,\end{cases} which has no solutions. We see that, with your solution, one obtains a linear system of 3 equations in 2 unknowns, which in general has no solution.
On the contrary, the solution with three terms leads to a linear system of $3$ equations in $3$ unknowns, which has a solution.
The general case is the following: if the fraction has in the denominator a multiple irreduucible factor $p(x)^r$, its contribution in the partial fractions decomposition is a sum: $$\frac{a_1(x)}{p(x)}+\frac{a_2(x)}{p(x)^2}+\dots+\frac{a_r(x)}{p(x)^r}, \quad\text{where}\quad \deg a_1(x), \dots, a_r(x)<\deg p(x).$$
$$\frac{2x^2 + 2x + 18}{x(x-3)^2} = \frac{A}{x} + \frac{Bx + C}{(x-3)^2} = \frac{A}{x} + \frac{Bx}{(x-3)^2} + \frac{C}{(x-3)^2} = \frac{A}{x} + \frac{B'}{x-3} + \frac{C}{(x-3)^2}$$