# How do you solve x^3-9x^2+14x=48 using the fundamental theorem of algebra?

Jul 14, 2016

The FTOA only tells you that it has $3$ roots, but we can find by other means that they are:

$8$ and $\frac{1}{2} \pm \frac{\sqrt{23}}{2} i$

#### Explanation:

We want to find the zeros of:

$f \left(x\right) = {x}^{3} - 9 {x}^{2} + 14 x - 48$

The fundamental theorem of algebra only tells us that since $f \left(x\right)$ is of degree $3$ then it has exactly $3$ Complex (possibly Real) zeros counting multiplicity. It does not tell us how to find them.

We can use the rational root theorem, which tells us that any rational zeros of $f \left(x\right)$ can be expressed in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 48$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 8 , \pm 12 , \pm 16 , \pm 24 , \pm 48$

Trying each in turn, we eventually find:

$f \left(8\right) = 512 - 576 + 112 - 48 = 0$

So $x = 8$ is a zero and $\left(x - 8\right)$ a factor:

${x}^{3} - 9 {x}^{2} + 14 x - 48 = \left(x - 8\right) \left({x}^{2} - x + 6\right)$

The remaining quadratic has negative discriminant, so only non-Real Complex zeros, but we can still use the quadratic formula to find them:

$x = \frac{1 \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(1\right) \left(6\right)}}{2 \cdot 1} = \frac{1}{2} \pm \frac{\sqrt{23}}{2} i$