Question

How do you use the rational roots theorem to find all possible zeros of `f(x) = 2x^3 + 3x^2 – 8x + 3`?

Answer 1

`f(x)` has zeros `1`, `1/2` and `-3`

Explanation:

`f(x) = 2x^3+3x^2-8x+3`

By the rational root theorem, 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 `3` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational zeros are:

`+-1/2, +-1, +-3/2, +-3`

We find:

`f(1) = 2+3-8+3 = 0`

So `x=1` is a zero and `(x-1)` a factor:

`2x^3+3x^2-8x+3 = (x-1)(2x^2+5x-3)`

Substituting `x=1/2` in the remaining quadratic we find:

`2x^2+5x-3 = 2(1/4)+5(1/2)-3 = 1/2+5/2-3 = 0`

So `x=1/2` is a zero and `(2x-1)` a factor:

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

So the final zero is `x=-3`