Question

What are all the possible rational zeros for `f(x)=2x^3-8x^2+15x-27` and how do you find all zeros?

Answer 1

Possible rational zeros:

`+-1/2, +-1, +-3/2, +-3, +-9/2, +-9, +-27/2, +-27`

Actual zeros:

`x = 3" "` and `" "x = 1/2(1+-sqrt(17)i)`

Explanation:

Given:

`f(x) = 2x^3-8x^2+15x-27`

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 `-27` 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, +-9/2, +-9, +-27/2, +-27`

Using Descartes' Rule of Signs, note that the pattern of the signs of the coefficients of `f(x)` is `+ - + -`. With `3` changes of sign, we can tell that `f(x)` has `3` or `1` positive real zeros. The pattern of the signs of the coefficients of `f(-x)` is `- - - -`. With no changes, that means that `f(x)` has no negative real zeros.

After a bit of trial and error, we find:

`f(3) = 2(color(blue)(3))^3-8(color(blue)(3))^2+15(color(blue)(3))-27`

`color(white)(f(3)) = 54-72+45-27 = 0`

So `x=3` is a zero and `(x-3)` a factor:

`2x^3-8x^2+15x-27 = (x-3)(2x^2-2x+9)`

Note that the discriminant of the remaining quadratic factor is negative, so it has non-real complex zeros.

We can complete the square (or use the quadratic formula) to find them.

For example:

`0 = 2(2x^2-2x+9)`

`color(white)(0) = 4x^2-4x+1+17`

`color(white)(0) = (2x-1)^2+(sqrt(17))^2`

`color(white)(0) = (2x-1)^2-(sqrt(17)i)^2`

`color(white)(0) = ((2x-1)-sqrt(17)i)((2x-1)+sqrt(17)i)`

`color(white)(0) = (2x-1-sqrt(17)i)(2x-1+sqrt(17)i)`

So:

`x = 1/2(1+-sqrt(17)i)`