Question

What are all the possible rational zeros for `f(x)=5x^4+32x^2-21`?

Answer 1

The "possible" rational zeros are:

`+-1/5, +-3/5, +-1, +-7/5, +-3, +-21/5, +-7, +-21`

The actual zeros are:

`+-sqrt(15)/5" "` and `" "+-sqrt(7)i`

Explanation:

Given:

`f(x) = 5x^4+32x^2-21`

If we apply the rational root theorem directly, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-21` and `q` a divisor of the coefficient `5` of the leading term.

That means that the only possible rational zeros are:

`+-1/5, +-3/5, +-1, +-7/5, +-3, +-21/5, +-7, +-21`

Note also that when `abs(x) >= 1`, then:

`f(x) >= 5+32-21 = 16`

So the only possible rational zeros are:

`+-1/5, +-3/5`

and since all the terms in `x` are of even degree, if `x_1` is a zero then so is `-x_1`.

So we only need to check `x=1/5` and `x=3/5`...

`f(1/5) = 5(1/625)+32(1/25)-21 = (1+160-2625)/125 = -2464/125`

`f(3/5) = 5(81/625)+32(9/25)-21 = (81+1440-2625)/125 = -1104/125`

So `f(x)` has no rational zeros.

`color(white)()`
We can factor `f(x)` as a quadratic in `x^2` using an AC method:

Look for a pair of factors of `AC=5*21 = 105` with difference `B=32`

The pair `35, 3` works.

Use this pair to split the middle term and factor by grouping:

`5x^4+32x^2-21 = (5x^4+35x^2)-(3x^2+21)`

`color(white)(5x^4+32x^2-21) = 5x^2(x^2+7)-3(x^2+7)`

`color(white)(5x^4+32x^2-21) = (5x^2-3)(x^2+7)`

Hence we can see that the zeros of `f(x)` are actually:

`x = +-sqrt(3/5) = +-sqrt(15)/5" "` and `" "x = +-sqrt(7)i`