Question

How do you use the rational roots theorem to find all possible zeros of `x^4-9x^2+20`?

Answer 1

Zeros: `x = +-2` and `x = +-sqrt(5)`

Explanation:

Given `f(x) = x^4-9x^2+20`, the rational roots theorem tells you that any rational zeros must be expressible in the form `p/q` for integers `p`, `q` with `p` a divisor of the constant term `20` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-4, +-5, +-10, +-20`

Evaluating `f(x)` for each in turn, this allows you to find the two rational zeros:

`f(2) = f(-2) = 16-36+20 = 0`

You can then divide `f(x)` by `(x-2)(x+2) = x^2-4` to get `x^2-5`, which has zeros `x = +-sqrt(5)`.

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Alternatively:

Note that `x^4-9x^2+20` is a quadratic in `x^2`. That is:

`x^4-9x^2+20 = (x^2)^2-9(x^2)+20`

Note also that `4+5 = 9` and `4xx5 = 20`

So we find:

`x^4-9x^2+20`

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

`=(x^2-2^2)(x^2-(sqrt(5))^2)`

`=(x-2)(x+2)(x-sqrt(5))(x+sqrt(5))`

Hence zeros:

`x = +-2` and `x = +-sqrt(5)`