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 \[\frac{x^2+x-6}{x^2+4\,x+3}\]

\({{x-2}\over{x+1}}\)

The polynomials in the numerator and denominator can be factorised by inspection: \[\begin{aligned} x^2+x-6 &= \left(x-2\right)\,\left(x+3\right)\\[0.25cm] x^2+4\,x+3 &= \left(x+1\right)\,\left(x+3\right)\end{aligned}\] So: \[\begin{aligned} \frac{x^2+x-6}{x^2+4\,x+3} &= \frac{\left(x-2\right)\,\left(x+3\right)}{\left(x+1\right)\,\left(x+3\right)}\\[0.25cm] &= {{x-2}\over{x+1}}\end{aligned}\] 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:

\[\begin{aligned} \mathrm{gcd}(x^2+x-6,x^2+4\,x+3) &= \mathrm{gcd}(x^2+4\,x+3,-3\,x-9)\\[0.1cm] {\small \blue{\text{because }x^2+x-6}} &\;{\small \blue{= (x^2+4\,x+3) -3\,x-9}}\\[0.4cm] &= \mathrm{gcd}(-3\,x-9,0)\\[0.1cm] {\small \blue{\text{because }x^2+4\,x+3}} &\;{\small \blue{= -{{x+1}\over{3}}\cdot (-3\,x-9) + 0}}\\[0.4cm] &= x+3\\[0.1cm] {\small \blue{\text{gcd (with leading coefficient 1)}}}& {\small \blue{\text{ read off from the previous result}}}\end{aligned}\] From \[\frac{x^2+x-6}{x+3}=x-2\quad\text{and}\quad \frac{x^2+4\,x+3}{x+3}=x+1\] follows that \[\frac{x^2+x-6}{x^2+4\,x+3}={{x-2}\over{x+1}}\]

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