Question

How do you use the rational roots theorem to find all possible zeros of `f(x)=2x^3-7x^2-10x-3`?

Answer 1

`f(x)` has zeros `-1/2` and `2+-sqrt(7)`

Explanation:

`f(x) = 2x^3-7x^2-10x-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`

We find:

`f(-1/2) = 2(-1/2)^3-7(-1/2)^2-10(-1/2)-3`

`=-1/4-7/4+10/2-3=0`

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

`2x^3-7x^2-10x-3`

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

`= (2x+1)(x^2-4x+4-7)`

`= (2x+1)((x-2)^2-(sqrt(7))^2)`

`= (2x+1)(x-2-sqrt(7))(x-2+sqrt(7))`

Hence the two other zeros are:

`x=2+-sqrt(7)`