Factorisation

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 $c$ .

Factorize $x^2 + 8 x+ 7$ .

We are looking for two numbers whose product equals $7$ and whose sum equals $8$ . The solution consists of two numbers $1$ and $7$ because then indeed $1\times 7 = 7$ and $1 + 7 = 8$ .

The requested factorisation is

$x^2 + 8 x+ 7 = (x + 1)(x + 7)\text.$

New example

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:

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 $\frac{p(x)}{q(x)}$ .