Limits part 2: Functions: Techniques
Factorisation
We saw that if we come across the indeterminate form \(\frac{0}{0}\) it often helps to factorise the numerator and denominator and divide them by common factors. In this paragraph we will discuss some methods to find such a factorisation.
First, we will discuss factorisation of quadratic polynomials. For any two numbers \(a\) and \(b\) it holds that \[ (x+a)(x+b) = x^2 + (a+b)x + ab\text. \] Therefore, we can write a quadratic polynomial \(x^2 + cx + d\) like \((x+a)(x+b)\) precisely if \(a + b = c\) and \(ab = d\). This means we are looking for two numbers that equal \(d\) when multiplied and add up to \(c\).
The requested factorisation is \[ x^2 + 6 x-16 = (x -2)(x + 8)\text. \]
If it is not easy to find such a decomposition you can use the quadratic formula to find the roots of the quadratic function. The fundamental theorem of algebra states:
Fundamental theorem of algebra Every polynomial \(p(x)=a_n x^n + a_{n-1}x^{n-1}+\dots+a_1 x + a_0\) can be written as \[ p(x) = a_n\left( x - x_1 \right)\left( x - x_2 \right)\cdots \left( x - x_n \right) \] where \(x_1, \dots, x_n\) are the roots of \(p\). It can happen that some roots appear more than once in the formula (the number of times is called the multiplicity of a zero).
Hence polynomials can be factorised into their zeros:
To factorise \(x^2 +4x +2\) we first calculate the roots using the quadratic formula. This gives \[ x = -2\pm\sqrt 2\text. \] This means that \[\begin{aligned}x^2 +4x +2 &= \left( x - \left( -2 + \sqrt 2\right) \right)\left( x - \left( -2 - \sqrt 2\right)\right)\\&= \left( x + 2 - \sqrt 2 \right)\left( x + 2 + \sqrt 2\right)\text. \end{aligned}\]
For polynomials higher degree it is not always possible to calculate the roots algebraically. You can then do a long division on the fraction \(\frac{p(x)}{q(x)}\).
