# Partial Fraction decomposition (degrees)

Let us say that I want to decompose the fraction

$$\frac{x-3}{x^2 +6x+5}$$ into partial fractions.

I know that we have to factor the denominator and write it as $$\frac{A}{x+1} + \frac{B}{x+5}.$$ Then we get

$$x - 3 = A(x+5) + B(x+1).$$

Select convenient values for $x$ and solve for $A$ and $B$.

My question, why if the denominator is linear, the numerator has to be 1 degree less than the denominator? Not just in this specific example, but in all questions - THE top has to be one degree less than the bottom. Why?

• For some basic information about writing math at this site see e.g. here, here, here and here. – GNUSupporter 8964民主女神 地下教會 Mar 19 '17 at 0:15
• when the polynomial in the numerator have less degree than the polynomial in the denominator you can find constants for $A$ and $B$, otherwise $A$ and/or $B$ will be polynomials. – Masacroso Mar 19 '17 at 0:26

The general theorem asserts that, given a rational fraction $\dfrac{N(x)}{D(x}$ in $K(x)$ ($K$ a field), with $\deg N(x)<\deg D(x)$, and a factorisation of $D(x)$ into a product of irreducible factors: $$D(x)=p_1(x)^{r_1}\dotsm p_n(x)^{r_n},$$ there is a decomposition $$\frac{N(x)}{D(x}=\frac{a_{1,1}(x)}{p_1(x)}+\dots+\frac{a_{1,r_1}(x)}{p_1(x)^{r_1}}+\dots+\frac{a_{n,1}(x)}{p_n(x)}+\dots+\frac{a_{n,r_n}(x)}{p_n(x)^{r_n}},$$ where $\,\deg a_k(x)<\deg p_k(x)$ for all $k=1,\dots n$.

Consider the fraction $\frac{p}{q}$ where $p$ and $q$ are polynomials, and the degree of $p$ is greater than or equal to the degree of $q$. Then we can use long division, to find a remainder of the form $\frac{r}{q}$ where the degree of $r$ is less than the degree of $q$.

Also with regards to why the numerator should be at least one degree less than the denominator, consider what would happen in your method if the numerator was the same degree or higher than the degree of the denominator.

If $f(x)=\frac{P(x)}{Q(x)}$ is a rational function and $\deg(P(x))\ge\deg(Q(x))$ we call $f$ an Improper rational function, it is said to be "top-heavy".

Conversely, if $\deg(P(x))\lt\deg(Q(x))$ then $f$ is a Proper rational function.

When $\deg(P(x))\ge\deg(Q(x))$, we use polynomial long division to turn a top-heavy rational function $f(x)=\frac{P(x)}{Q(x)}$ into a proper rational function:

$$f(x)=S(x) + \frac {R(x)}{Q(x)}$$

where $S(x)$ and $R(x)$ are polynomials and $\deg(R(x))\lt\deg(Q(x))$.

Select convenient values for $x$ and solve for $A$ and $B$.

Finding $A$ and $B$ (and often $C$, $D$, $E$ and so on) is extremely important when you get to integral calculus if you need to evaluate integrals involving rational functions such as this one using the function from your example:

$$\int \frac{x-3}{x^2 +6x+5} \, dx$$ .