Polynomial Roots

The roots (sometimes called zeroes or solutions) of a polynomial $P(x)$ are the values of $x$ for which $P(x)$ is equal to zero. Finding the roots of a polynomial is sometimes called solving the polynomial.

For example, if $P(x)=x^2-5x+6$, then the roots of the polynomial $P(x)$ are $2$ and $3$, since both $P(2)$ and $P(3)$ are equal to zero.

A polynomial is a special kind of mathematical expression that looks like this:

$$a_n x^n+a_{n-1}x^{n-1}+a_{n-2}+x^{n-2}+\cdots+a_2x^2+a_1x+a_0=\displaystyle\sum_{i=0}^n a_i x^i.$$

If $a_n$ is not equal to zero, then we say that the polynomial has degree $n$. According to the fundamental theorem of algebra any polynomial with degree $n$ has $n$ complex roots, counted with multiplicity.

Finding the root of a linear polynomial (a polynomial with degree one) $ax+b$ is very straightforward. The formula for the root is $-\frac{b}{a}$ (although calling this a formula is going a bit overboard).

The roots for a quadratic polynomial (a polynomial with degree two) $ax^2+bx+c$ is given by the formula $$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

The formula for the roots of a cubic polynomial (a polynomial with degree three) is a bit more complicated while the formula for the roots of a quartic polynomial (a polynomial with degree four) would fill two blackboards!

What about fifth-degree (quintic) polynomials? What about polynomials with higher degrees?

About $170$ years ago, a young mathematician by the name of Henrik Abel proved that it is impossible to find a formula for the solutions of a quintic polynomial by adding, subtracting, multiplying, dividing and taking $n^\text{th}$ roots. More formally speaking, a quintic polynomial is not solvable by radicals.

Henrik Abel

The same is true for polynomials with higher degrees.

$1$. If the coefficients of a polynomial are real and if $a+ib$ is a root of that polynomial, then so is $a-ib$. See Complex Conjugate Root Theorem.

$2$. If a polynomial with rational coefficients has $a + \sqrt{b}$ as a root, where $a, b$ are rational and $\sqrt{b}$ is irrational, then $a - \sqrt{b}$ is also a root.

$1$. Polynomials

$2$. Vieta's Formula

$3$. Factoring Polynomials