Question

How do you use the rational roots theorem to find all possible zeros of `P(x) = x^5 -2x^4 +4x -8`?

Answer 1

The only Real zero is `x=2`

Explanation:

By the rational roots theorem, any rational zeros of `P(x)` will be expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-8` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1`, `+-2`, `+-4`, `+-8`

Trying each in turn, we find:

`P(2) = 32-32+8-8 = 0`

So `x=2` is a zero and `(x-2)` a factor of `P(x)`:

`x^5-2x^4+4x-8 = (x-2)(x^4+4)`

Note that `x^4 >= 0` for all Real values of `x` so `x^4+4 >= 4` has no Real zeros.

It does have Complex zeros `x=1+-i` and `x=-1+-i`, being the `4`th roots of `-4`.

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Bonus

Note that although `x^4+4` has no Real zeros or linear factors, it is possible to factor it into a product of two quadratics with Real coefficients.

In fact, we find:

`x^4+4 = (x^2-2x+2)(x^2+2x+2)`

This is an instance of a nice identity for factoring quartics:

`(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4`

Note in particular that if we put `k=sqrt(2)` we find:

`(a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2) = a^4+b^4`

In our case we used this with `k=sqrt(2)`, `a=x` and `b=sqrt(2)`