# 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:

don't confuse

*a*with αdon't confuse

*b*with βdon't confuse

*g*with γdon't confuse

*d*with δ

a,b,c,d&eare real constants

α, β, γ, δare roots of polynomials

## Quadratic Polynomials

**A quadratic equation in the form ***a***x² + ***b***x + ***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:

## Cubic Polynomials

**A cubic equation in the form ***a***x³ + ***b***x² + ***c***x + ***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:

## Quartic Polynomials

**A quartic equation in the form ***a***x⁴ + ***b***x³ + ***c***x² + ***d***x + ***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:

## Linear Transformations of Roots

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:

Use the roots of polynomial equations above

Rearrange the roots and substitute into the first equation

### Method 1

**(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

**Method 2**

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**

## Comments