Factorising polynomials

A polynomial can be factorised into a product of polynomials. Suppose we have a polynomial

$$P_n(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\dots +a_1{x}^1+a_0{x}^0,$$

The constant n is the degree of the polynomial, that is, the highest index that appears in the polynomial with a non-zero coefficient. For example, the polynomial x³ − 4x² − 5x has degree 3. All the other indices are either lower (the indices 2 and 1) or have a zero coefficient. Now we can say that if we find a root k of our polynomial, then there is another polynomial Q_{n − 1}(x), of degree one lower, such that

$$P_n(x)=(x-k)\cdot Q_{n-1}(x).$$

That sounds complicated, but what it says is that if we know a root of the polynomial P_n(x), we can find another polynomial Q_{n − 1}(x) which, when multiplied by (x − k), gives the original polynomial P_n(x), and Q_{n − 1}(x) has degree one less than P_n(x). Looking at our polynomial P₃(x) = x³ − 4x² − 5x, we can see that one of its roots is k₁ = 0 (because if we substitute zero for x, the whole expression is zero). So there must be a polynomial Q₂(x) such that

$$P_3(x)=(x-0)\cdot Q_2(x)$$

This is easy, because x − 0 is of course just x, so we get the equation

$$P_3(x)=x\cdot Q_2(x)$$

We just need to take out the common factor x from the original polynomial P₃(x) to get our new polynomial Q₂(x):

$$P_3(x)=x\cdot (x^2-4x-5)$$

That is the first factorisation of the polynomial into a product. We have factorised the original polynomial x³ − 4x² − 5x as (x − 0) times x² − 4x − 5, which is a polynomial of lower degree. But we can factorise x² − 4x − 5 further. If we solve the quadratic equation

$$x^2-4x-5=0$$

we find that its roots are k₂ = 5 and k₃ = −1. So we should be able to find a polynomial Q₁(x) such that

$$Q_2(x)=(x-5)\cdot Q_1(x)$$

Finding Q₁(x) is not entirely straightforward, but we can use algebraic division to help. That tells us that the resulting polynomial is Q₁(x) = x + 1, so we can write

$$x^2-4x-5=(x-5)\cdot (x+1)$$

The whole original polynomial P₃(x) is therefore equal to:

$$P_3(x)=x\cdot (x-5)\cdot (x+1)$$

Once we have factorised the whole polynomial like this, we can read off the roots easily. If any one of the factors in the product is zero, the whole expression is zero. So it is clear from this form that the roots of the polynomial are exactly the numbers k₁ = 0, k₂ = 5 and k₃ = −1.