Polynomial Decomposition Theorem

The fundamental theorem of algebra for polynomial equations guarantees that "every degree polynomial n≥ 1 has at least one complex root". The proof of this theorem was made by the mathematician Friedrich Gauss in 1799. From it, we can demonstrate the polynomial decomposition theorem, which guarantees that any polynomial can be decomposed into first-degree factors. Take the following polynomial p(x) of grade n ≥ 1 and the_no ≠ 0:

p(x) = a_no x^no + the_n-1 x^n-1 + … + the₁x¹ + the₀

Through the fundamental theorem of algebra, we can state that this polynomial has at least one complex root. u₁, such that p(u₁) = 0. O D'Alembert's theorem to the division of polynomials states that if p(u₁) = 0, then p(x) is divisible by (x - u₁), resulting in a quotient what₁(x), which is a degree polynomial (n - 1), which leads us to say:

p (x) = (x - u₁). what₁(x)

From this equation, it is necessary to highlight two possibilities:

If u = 1 and what₁(x) is a polynomial of degree (n - 1), then what₁(x) has degree 0. As the dominant coefficient of

p(x) é The_no, what₁(x) is a constant polynomial of type what₁(x)=The_no. So we have:

p (x) = (x - u₁). what₁(x)
(x) = (x - u₁). The_no
p(x) = a_no . (x - u₁)

But if u ≥ 2, then the polynomial what₁ has degree n - 1 ≥ 1 and the fundamental theorem of algebra holds. We can say that the polynomial what₁ has at least one root no₂, which leads us to say that what₁ can be written as:

what₁(x) = (x - u₂). what₂(x)

But how p (x) = (x - u₁). what₁(x), we can rewrite it as:

p (x) = (x - u₁). (x - u₂). what₂(x)

Successively repeating this process, we will have:

p(x) = a_no. (x - u₁). (x - u₂) … (x – u_no)

Thus, we can conclude that every polynomial or polynomial equation p(x) = 0 of grade n≥ 1 own exactly no complex roots.​

Example: Be p(x) a polynomial of degree 5, such that its roots are – 1, 2, 3, – 2 and 4. Write this polynomial decomposed into 1st degree factors, considering the dominant coefficient equal to 1. It must be written in extended form:

if – 1, 2, 3, – 2 and 4 are roots of the polynomial, so the product of the differences of x for each of these roots results in p(x):

p(x) = a_no.(x + 1).(x – 2).(x – 3).(x + 2).(x – 4)

If the dominant coefficient The_no = 1, we have:

p (x) = 1.(x + 1).(x – 2).(x – 3).(x + 2).(x – 4)
p (x) = (x + 1).(x – 2).(x – 3).(x + 2).(x – 4)
p (x) = (x² - x - 2).(x - 3).(x + 2).(x - 4)
p (x) = (x³ – 4x² + x + 6).(x + 2).(x – 4)
p(x) = (x⁴ – 2x³ – 7x² + 8x + 12). (x – 4)
p(x) = x⁵ – 6x⁴ + x³ + 36x² - 20x - 48

By Amanda Gonçalves
Graduated in Mathematics

Source: Brazil School - https://brasilescola.uol.com.br/matematica/teorema-decomposicao-um-polinomio.htm