Decomposing polynomials into products

A polynomial, or polynomial, can be decomposed into a product of polynomials. Let's imagine we have a polynomial

$$P_n(x)=a_nx^n + a_{n-1}x^{n-1}+\ldots+a_1x^1+a_0x^0,$$

The constant n denotes the degree of the polynomial, that is, the highest exponent found in the polynomial that does not have a zero coefficient. For example, the polynomial x³ − 4x² − 5x would have degree 3. All other exponents are either lower (exponents 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), which is a degree less and

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

This sounds complicated, but effectively it says that if we know the root of the polynomial P_n(x), we are able to find another polynomial Q_{n − 1}(x), which when multiplied by the expression (x − k), we get the original polynomial P_n(x), with the polynomial Q_{n − 1}(x) being one degree smaller than the polynomial P_n(x). When we look at our polynomial P₃(x) = x³ − 4x² − 5x, we see that one of the roots of the polynomial is the number k₁ = 0 (because if we substitute zero after x in the polynomial, the whole expression is equal to zero). And so it must be true that there exists a polynomial Q₂(x) such that

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

Here we'll keep it simple, because x − 0 is obviously equal to x and so we get the equation:

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

From the original polynomial P₃(x) we just need to extract x and we get our new polynomial Q₂(x):

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

This is the first decomposition of the polynomial into a product. We were able to decompose the original polynomial x³ − 4x² − 5x into the product (x − 0) times x² − 4x − 5, which is a polynomial of lower degree. But we can again decompose this polynomial x² − 4x − 5 further. If we solve the quadratic equation

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

we find that the roots of this equation are the numbers k₂ = 5 and k₃ = −1. We should therefore be able to find a polynomial Q₁(x) such that

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

Finding the polynomial Q₁(x) is not quite so easy, we can help ourselves by dividing polynomials for example. This will show that the resulting polynomial is Q₁(x) = x + 1, so we can write

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

The full original polynomial P₃(x) is equal to:

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

If we decompose the whole polynomial in this way, we are able to read the individual roots nicely. If any of the terms in the product is equal to zero, the whole expression will be equal to zero. Therefore, it is clear from the notation that the roots of the polynomial are just the numbers k₁ = 0, k₂ = 5 and k₃ = −1.