Question

What are all the possible rational zeros for `f(x)=2x^4-6x^2+5x-15` and how do you find all zeros?

Answer 1

"Possible" rational zeros are:

`+-1/2, +-1, +-3/2, +-5/2, +-3, +-5, +-15/2, +-15`

For actual zeros, read on...

Explanation:

Given:

`f(x) = 2x^4-6x^2+5x-15`

By the rational roots 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 `-15` 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, +-5/2, +-3, +-5, +-15/2, +-15`

With the function as specified, none of these is a zero.

If we graph `f(x)` then it looks like this:

graph{2x^4-6x^2+5x-15 [-5, 5, -35, 25]}

There are two irrational zeros, one somewhere in the interval `(-5/2, -2)`, the other in the interval `(3/2, 2)`.

Typically for a quartic, these are possible but somewhat messy to find algebraically. If you would really like to see a solution, please request in the comments.

I suspect a typo in the question.

Consider the function:

`f_1(x) = 2x^3-6x^2+5x-15`

The rational roots theorem suggests the same possible rational roots, but this time the graph looks like this:

graph{2x^3-6x^2+5x-15 [-5, 5, -35, 25]}

There is one real rational root at `x=3`:

`f_1(3) = 2(color(blue)(3))^3-6(color(blue)(3))^2+5(color(blue)(3))-15`

`color(white)(f_1(3)) = 54-54+15-15`

`color(white)(f_1(3)) = 0`

We can also factor `f_1(x)` by grouping:

`2x^3-6x^2+5x-15 = (2x^3-6x^2)+(5x-15)`

`color(white)(2x^3-6x^2+5x-15) = 2x^2(x-3)+5(x-3)`

`color(white)(2x^3-6x^2+5x-15) = (2x^2+5)(x-3)`

From this we can see the real zero `x=3` and find the complex zeros:

`x = +-sqrt(5/2)i = +-sqrt(10/4)i = +-sqrt(10)/2i`