# Roots and zeros

When we solve polynomial equations with degrees greater than zero, it may have one or more real roots or one or more imaginary roots. In mathematics, the fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root.

Further on every non-zero single-variable polynomial with complex coefficients has exactly as many complex roots as its degree, if each root is counted up to its multiplicity.

If a+bi is a zero (root) then a-bi is also a zero of the function.

**Example**

Show that if \((2+i)\) is a zero to \(f(x)=-x^{2}+4x-5\) then \(2-i\) is also a zero of the function(this example is also shown in our video lesson).

First we check if \((2+i)\) is a zero to \(f(x)\) by plugging the zero into our function:

$$f(x)=-x^{2}+4x-5$$

$$f(2+i)=-(2+i)^{2}+4(2+i)-5=$$

$$=-(4+i^{2}+4i)+8+4i-5=$$

$$=-(4-1+4i)+3+4i=$$

$$=-3-4i+3+4i=0$$

(2+i) is a zero now (2-i) also must be a zero; we control this by plugging (2-i) into our function:

$$f(x)=-x^{2}+4x-5$$

$$f(2-i)=-(2-i)^{2}+4(2-i)-5=$$

$$=-(4+i^{2}-4i)+8-4i-5=$$

$$=-(4-1-4i)+3-4i=$$

$$=-3+4i+3-4i=0$$

**Video lesson**

the previous example.