# How do you find the number of roots for f(x) = x^3 – 75x + 250 using the fundamental theorem of algebra?

Jun 18, 2016

The FTOA tells us that there are $3$ zeros counting multiplicity.
Further investigation shows us that they are $x = 5$ with multiplcity $2$ and $x = - 10$.

#### Explanation:

The fundamental theorem of algebra tells you that any non-constant polynomial in one variable with Complex (possibly Real) coefficients has a zero which is a Complex (possibly Real) number.

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

So in our example:

$f \left(x\right) = {x}^{3} - 75 x + 250$

is of degree $3$, so has exactly $3$ zeros counting multiplicity.

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Bonus

What else can we find out about these zeros?

By the rational root theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $250$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are positive and negative factors of $250$:

$\pm 1$, $\pm 2$, $\pm 5$, $\pm 10$, $\pm 25$, $\pm 50$, $\pm 125$, $\pm 250$

We find:

$f \left(- 10\right) = - 1000 + 750 + 250 = 0$

So $x = - 10$ is a zero and $x + 10$ a factor:

${x}^{3} - 75 x + 250$

$= \left(x + 10\right) \left({x}^{2} - 10 x + 25\right)$

$= \left(x + 10\right) \left(x - 5\right) \left(x - 5\right)$

So the zeros are $x = - 10$ and $x = 5$ with multiplicity $2$.