Question

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

Answer 1

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(x) = x^3-75x+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(x)` must be expressible in the form `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`:

`+-1`, `+-2`, `+-5`, `+-10`, `+-25`, `+-50`, `+-125`, `+-250`

We find:

`f(-10) = -1000+750+250 = 0`

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

`x^3-75x+250`

`=(x+10)(x^2-10x+25)`

`=(x+10)(x-5)(x-5)`

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