# How do you find the number of roots for f(x) = 3x^4 + x + 2 using the fundamental theorem of algebra?

Dec 7, 2016

By the FTOA, $f \left(x\right)$ has exactly $4$ Complex, possibly Real zeros counting multiplicity.

Further we find that none are Real.

#### Explanation:

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

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

In our example:

$f \left(x\right) = 3 {x}^{4} + x + 2$

is of degree $4$ and therefore has $4$ Complex, possibly Real, zeros counting multiplicity.

What else can we find out about these zeros?

The pattern of signs of the coefficients of $f \left(x\right)$ is $+ + +$. With no changes of sign, we can deduce that $f \left(x\right)$ has no positive Real zeros by Descartes' Rule of Signs.

The pattern of signs of $f \left(- x\right)$ is $+ - +$. With $2$ changes of sign, it means that $f \left(x\right)$ has $0$ or $2$ negative Real zeros.

Note that when $\left\mid x \right\mid \le 1$, we have $3 {x}^{4} \ge 0$ and since $\left\mid x \right\mid < 2$, we find:

$f \left(x\right) > 0$

When $\left\mid x \right\mid > 1$, then $\left\mid 3 {x}^{4} \right\mid = \left\mid 3 {x}^{3} \cdot x \right\mid > 3 \left\mid x \right\mid$
Hence:

$f \left(x\right) > 0$

So $f \left(x\right)$ has no Real zeros. All of its $4$ zeros must be non-Real Complex.