# How do you find the number of roots for 0=x^3-4x-48 using the fundamental theorem of algebra?

Dec 9, 2015

The degree of this polynomial is $3$ so by the fundamental theorem of algebra it has $3$ roots (counting multiplicity).

#### Explanation:

A non-zero polynomial equation in one variable always has as many roots as its degree. They may be repeated and/or Complex, but it will have that many roots.

The highest degree term in $0 = {x}^{3} - 4 x - 48$ is the ${x}^{3}$ one, so the polynomial is of degree $3$.

Let $f \left(x\right) = {x}^{3} - 4 x - 48$.

Notice that $f \left(4\right) = 64 - 16 - 48 = 0$, so $x = 4$ is one of the roots and $\left(x - 4\right)$ is a factor of $f \left(x\right)$.

${x}^{3} - 4 x - 48 = \left(x - 4\right) \left({x}^{2} + 4 x + 12\right)$

The remaining quadratic factor $\left({x}^{2} + 4 x + 12\right)$ is of the form $a {x}^{2} + b x + c$, with $a = 1$, $b = 4$ and $c = 12$. It has zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- 4 \pm \sqrt{{\left(- 4\right)}^{2} - \left(4 \times 1 \times 12\right)}}{2 \times 1}$

$= \frac{- 4 \pm \sqrt{- 32}}{2} = - 2 \pm 2 \sqrt{2} i$

..that is a conjugate pair of Complex roots.