Normal form of a rational function

Basic functions: Rational functions

Normal form of a rational function

Just as fractions can be uniquely written in an irrecudible form, rational functions can also be written in a unique irreducible form.

Each rational function has a unique form

$\dfrac{p(x)}{q(x)}$ where $p(x)$ and $q(x)$ are mutually indivisible polynomials, i.e., with $\gcd\left(p(x),q(x)\right)=1$ , and $q(x)$ has leading coefficient $1$ .

Such a fraction is called irreducible. We also call it the normal form of a rational function.

Analytical continuation

$\frac{4(x-1)^2}{2x^2-2}=\frac{4(x-1)^2}{2(x-1)(x+1)}=\frac{2(x-1)}{(x+1)}$ As you can see, the perforation $x=1$ has been removed from the original rational function $\frac{4(x-1)^2}{2x^2-2}$ . Converting the rational function into the normal form $\frac{2(x-1)}{(x+1)}$ is thus accompanied by an extension of the domain of the function via an analytical continuation of the original rational function.

Determine the normal form of

$\frac{x^2+11\,x+30}{x^2+4\,x-5}$

${{x+6}\over{x-1}}$

The polynomials in the numerator and denominator can be factorised by inspection:

$\begin{aligned} x^2+11\,x+30 &= \left(x+5\right)\,\left(x+6\right)\\[0.25cm] x^2+4\,x-5 &= \left(x-1\right)\,\left(x+5\right)\end{aligned}$ So:

$\begin{aligned} \frac{x^2+11\,x+30}{x^2+4\,x-5} &= \frac{\left(x+5\right)\,\left(x+6\right)}{\left(x-1\right)\,\left(x+5\right)}\\[0.25cm] &= {{x+6}\over{x-1}}\end{aligned}$ Because the leading coefficient of the denominator in the previous expression is equal to $1$ , we have indeed obtained the normal form.

$\phantom{abc}$

Of course, instead of factorisation, we could have found the greatest common divisor of the numerator and denominator, and then divided both the numerator and denominator by this:

$\begin{aligned} \mathrm{gcd}(x^2+11\,x+30,x^2+4\,x-5) &= \mathrm{gcd}(x^2+4\,x-5,7\,x+35)\\[0.1cm] {\small \blue{\text{because }x^2+11\,x+30}} &\;{\small \blue{= (x^2+4\,x-5) + 7\,x+35}}\\[0.4cm] &= \mathrm{gcd}(7\,x+35,0)\\[0.1cm] {\small \blue{\text{because }x^2+4\,x-5}} &\;{\small \blue{= {{x-1}\over{7}}\cdot (7\,x+35) + 0}}\\[0.4cm] &= x+5\\[0.1cm] {\small \blue{\text{gcd (with leading coefficient 1)}}}& {\small \blue{\text{ read off from the previous result}}}\end{aligned}$ From

$\frac{x^2+11\,x+30}{x+5}=x+6\quad\text{and}\quad \frac{x^2+4\,x-5}{x+5}=x-1$ follows that

$\frac{x^2+11\,x+30}{x^2+4\,x-5}={{x+6}\over{x-1}}$

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