Roots of Polynomials

Won't lie, this topic isn't particularly fun. It's just a whole lot of algebra and equations that you need to know and get right. At least the topic is small.

To be perfectly honest, the only challenge is writing everything out clearly, and not getting confused between the Greek and Latin alphabets:

a, b, c, d & e are real constants

α, β, γ, δ are roots of polynomials

A quadratic equation in the form ax² + bx + c = 0 has two possible roots, α and β. These can be distinct and real, repeated (the same), or complex conjugates.

The roots can be added or multiplied to give the following results:

Squaring or cubing the roots gives different results:

A cubic equation in the form ax³ + bx² + cx + d = 0 has three possible roots, α, β and γ.

The roots can be added or multiplied to give the following results:

Squaring or cubing the roots gives different results:

A quartic equation in the form ax⁴ + bx³ + cx² + dx + e = 0 has four possible roots, α, β, γ and δ.

The roots can be added or multiplied to give the following results:

Squaring the roots gives different results:

If you know the sums and products of roots of a polynomial, you can find the equation of another polynomial whose roots are a linear transformation of the first.

For example, if the equation x² - 2x + 3 = 0 has roots α and β, find the equation that has roots (α+2) and (β+2).

This can be done in two ways:

(α+2) + (β+2) = α + β + 4 = -b/a + 4 = 6

(α+2)(β+2) = αβ + 2α + 2β + 4 = αβ + 2(α+β) + 4 = c/a - 2b/a + 4 = 11

So the equation is w² - 6w + 11 = 0

The coefficient of each term must be the same as the initial equation

Let w = x + 2

Rearrange for x: x = w - 2

Substitute into first equation: (w-2)² - 2(w-2) + 3 = 0

w² - 4w + 4 -2w + 4 +3 = 0

w² - 6w + 11 = 0