# Roots

A root of a polynomial is a solution to the equation where the polynomial is set equal to zero.

The fundamental theorem of algebra states that for a polynomial in one variable, the number of roots is equal to the degree of the polynomial (though some may be double or multiple roots).

Example 1:

Find the roots of the polynomial ${x}^{2}-5x+6$ .

Equate the polynomial to zero.

${x}^{2}-5x+6=0$

In this case, the polynomial can be easily factored :

$\left(x-2\right)\left(x-3\right)=0$

By the zero product property , either $x=2$ or $x=3$ .

(This polynomial has degree $2$ , so we have found the $2$ roots.)

In the above example, both roots are positive integers. In other polynomials, roots may involve radicals and/or complex numbers .

Example 2:

Find the roots of the polynomial $2{x}^{3}+2{x}^{2}+3x$ .

Notice that we can immediately factor out an $x$ .

$x\left(2{x}^{2}+2x+3\right)=0$

By the zero product property , either $x=0$ or $2{x}^{2}+2x+3=0$ .

So, one root is $0$ . To find the other two roots, we use the quadratic formula :

$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

Here $a=2,b=2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=3$ .

$x=\frac{-2±\sqrt{{2}^{2}-4\left(2\right)\left(3\right)}}{2\left(2\right)}$

Simplify.

$x=\frac{-2±\sqrt{-20}}{4}$

$x=\frac{-2±2i\sqrt{5}}{4}$

$x=-\frac{1}{2}±i\frac{\sqrt{5}}{2}$

So the polynomial has $1$ real root and $2$ complex roots, for a total of $3$ .