Question

How do you use the rational roots theorem to find all possible zeros of `P(x) = x^5 + 3x^3 + 2x - 6`?

Answer 1

The "possible" rational zeros are `+-1, +-2, +-3, +-6`.

The only real zero is `x=1`

Explanation:

Given:

`P(x) = x^5+3x^3+2x-6`

By the rational roots theorem any rational zeros of `P(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-6` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-3, +-6`

In addition, note that the sum of the coefficients is zero, i.e.:

`1+3+2-6 = 0`

Hence we can deduce that `x=1` is a zero and `(x-1)` a factor:

`x^5+3x^3+2x-6 = (x-1)(x^4+x^3+4x^2+4x+6)`

The remaining quartic has all positive coefficients, so any real zeros must be negative.

Trying `-1`, `-2`, `-3` and `-6` we find that none is a zero.

In fact it has only non-real complex zeros.