# How do you find the number of roots for 0=x^3+450x^2-1350 using the fundamental theorem of algebra?

Dec 28, 2015

This polynomial equation is of degree $3$, so has exactly $3$ Complex roots counting multiplicity, some of which may be Real.

Actually all three roots are Real and distinct.

#### Explanation:

The fundamental theorem of algebra tells you that a polynomial equation in one variable of degree $n > 0$ has at least one Complex root.

A simple corollary of this is that a polynomial equation in one variable of degree $n > 0$ has $n$ possibly repeated Complex roots. This is sometimes included as part of the statement of the fundamental theorem of algebra.

Here's a sketch of the proof of the corollary:

If a polynomial $f \left(x\right)$ has one Complex zero ${x}_{1}$, then $\left(x - {x}_{1}\right)$ is a factor of $f \left(x\right)$ and you can write $f \left(x\right) = \left(x - {x}_{1}\right) g \left(x\right)$ where $g \left(x\right)$ is of degree $n - 1$. Then if $n - 1 > 0$, $g \left(x\right) = 0$ has at least one Complex root ${x}_{2}$, which might be the same as ${x}_{1}$, etc. So counting multiplicity, a polynomial of degree $n$ has exactly $n$ Complex roots.

Note that the Complex roots referred to in all of this may be Real or non-Real.

The equation $0 = {x}^{3} + 450 {x}^{2} - 1350$ is of degree $3$, hence has $3$ Complex roots counting multiplicity. That's all the fundamental theorem of algebra will tell you.

Beyond this, since this polynomial has Real coefficients, any non-Real Complex roots will occur in conjugate pairs. So we can deduce that our polynomial will have at least one Real root.

It so happens that all of these $3$ roots are Real in this example, being approximately $- 450$, $- 1.74$ and $1.73$

To show that all of the roots are Real, let $f \left(x\right) = {x}^{3} + 450 {x}^{2} - 1350$ and evaluate $f \left(x\right)$ for a few values:

$f \left(- 450\right) = - {450}^{3} + {450}^{3} - 1350 = - 1350 < 0$

$f \left(- 2\right) = - 8 + 450 \cdot 4 - 1350 = - 8 + 1800 - 1350 = 442 > 0$

$f \left(0\right) = 0 + 0 - 1350 = - 1350 < 0$

$f \left(2\right) = 8 + 450 \cdot 4 - 1350 = 8 + 1800 - 1350 = 458 > 0$

So $f \left(x\right)$ crosses the $x$ axis $3$ times and $f \left(x\right) = 0$ has $3$ distinct Real roots.