# How do you find the number of roots for f(x) = 2x^3 – 12x^2 + 11x + 2 using the fundamental theorem of algebra?

Nov 25, 2016

The fundamental theorem of algebra tells us that this cubic has $3$ zeros counting multiplicity.

We can further discover that all $3$ are Real with $2$ distinct positive zeros and one negative.

#### Explanation:

$f \left(x\right) = 2 {x}^{3} - 12 {x}^{2} + 11 x + 2$

The fundamental theorem of algebra (FTOA) tells us that any polynomial of degree $n > 0$ has at least one Complex, possibly Real, zero.

A straightforward corollary of that, often stated as part of the FTOA is that a polynomial of degree $n > 0$ will have exactly $n$ Complex (possibly Real) zeros counting multiplicity.

This follows since given a polynomial of degree $n > 0$ with a zero $x = a$, we can divide the polynomial by $\left(x - a\right)$ to give a polynomial of degree $n - 1$. If $n - 1 > 0$ then it has a possibly distinct zero $b$, etc.

In our example, $f \left(x\right)$ is of degree $3$ so has exactly $3$ Complex, possibly Real, zeros counting multiplicity.

We can find out more by examining the discriminant $\Delta$, which for a cubic of the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

In our example, $a = 2$, $b = - 12$, $c = 11$ and $d = 2$, so we find:

$\Delta = 17424 - 10648 + 13824 - 432 - 9504 = 10664$

Since $\Delta > 0$, all three zeros of $f \left(x\right)$ are Real and distinct.

Using Descartes's Rule of Signs we can also tell that $2$ of these zeros are positive and one negative.