# How do you find the number of roots for f(x) = 4x^3 – x^2 – 2x + 1 using the fundamental theorem of algebra?

Jul 29, 2016

It has $3$ zeros since it is of degree $3$.

#### Explanation:

The fundamental theorem of algebra (FTOA) tells you that a polynomial of non-zero degree has at least one Complex (possibly Real) zero.

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

To see this, note that if $f \left(x\right)$ is of degree $n > 0$, then by the FTOA it has a zero ${r}_{1} \in \mathbb{C}$. Then $\left(x - {r}_{1}\right)$ is a factor of $f \left(x\right)$ and $f \frac{x}{x - {r}_{1}}$ is a polynomial of degree $n - 1$. If $n - 1 > 0$ then $f \frac{x}{x - {r}_{1}}$ has a zero ${r}_{2} \in \mathbb{C}$ (possibly equal to ${r}_{1}$), so a factor $\left(x - {r}_{2}\right)$, etc. Repeat to find all $n$ zeros.

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

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Bonus

The discriminant $\Delta$ of a cubic polynomial in 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 this example, $a = 4$, $b = - 1$, $c = - 2$, $d = 1$ and we find:

$\Delta = 4 + 128 + 4 - 432 + 144 = - 152 < 0$

Since $\Delta < 0$ this cubic has one Real zero and a Complex conjugate pair of non-Real zeros.

We can use Descartes' rule of signs to tell that $f \left(x\right)$ has $0$ or $2$ positive Real zeros and exactly one $1$ negative zero. Since we already know that it only has one Real zero, it is negative.