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 \[\frac{x^2+x-30}{x^2-2\,x-15}\]

\({{x+6}\over{x+3}}\)

The polynomials in the numerator and denominator can be factorised by inspection: \[\begin{aligned} x^2+x-30 &= \left(x-5\right)\,\left(x+6\right)\\[0.25cm] x^2-2\,x-15 &= \left(x-5\right)\,\left(x+3\right)\end{aligned}\] So: \[\begin{aligned} \frac{x^2+x-30}{x^2-2\,x-15} &= \frac{\left(x-5\right)\,\left(x+6\right)}{\left(x-5\right)\,\left(x+3\right)}\\[0.25cm] &= {{x+6}\over{x+3}}\end{aligned}\] 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:

\[\begin{aligned} \mathrm{gcd}(x^2+x-30,x^2-2\,x-15) &= \mathrm{gcd}(x^2-2\,x-15,3\,x-15)\\[0.1cm] {\small \blue{\text{because }x^2+x-30}} &\;{\small \blue{= (x^2-2\,x-15) + 3\,x-15}}\\[0.4cm] &= \mathrm{gcd}(3\,x-15,0)\\[0.1cm] {\small \blue{\text{because }x^2-2\,x-15}} &\;{\small \blue{= {{x+3}\over{3}}\cdot (3\,x-15) + 0}}\\[0.4cm] &= x-5\\[0.1cm] {\small \blue{\text{gcd (with leading coefficient 1)}}}& {\small \blue{\text{ read off from the previous result}}}\end{aligned}\] From \[\frac{x^2+x-30}{x-5}=x+6\quad\text{and}\quad \frac{x^2-2\,x-15}{x-5}=x+3\] follows that \[\frac{x^2+x-30}{x^2-2\,x-15}={{x+6}\over{x+3}}\]

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