Question

How do you solve `x^3-9x^2+14x=48` using the fundamental theorem of algebra?

Answer 1

The FTOA only tells you that it has `3` roots, but we can find by other means that they are:

`8` and `1/2+-sqrt(23)/2i`

Explanation:

We want to find the zeros of:

`f(x) = x^3-9x^2+14x-48`

The fundamental theorem of algebra only tells us that since `f(x)` is of degree `3` then it has exactly `3` Complex (possibly Real) zeros counting multiplicity. It does not tell us how to find them.

We can use the rational root theorem, which tells us that any rational zeros of `f(x)` can be expressed in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-48` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-3, +-4, +-6, +-8, +-12, +-16, +-24, +-48`

Trying each in turn, we eventually find:

`f(8) = 512-576+112-48 = 0`

So `x=8` is a zero and `(x-8)` a factor:

`x^3-9x^2+14x-48 = (x-8)(x^2-x+6)`

The remaining quadratic has negative discriminant, so only non-Real Complex zeros, but we can still use the quadratic formula to find them:

`x = (1+-sqrt((-1)^2-4(1)(6)))/(2*1) = 1/2+-sqrt(23)/2i`