Question

How do you find all the zeros of `f(x)= 2x^3 + 3x^2+ 8x- 5`?

Answer 1

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

Explanation:

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

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 `-5` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible ratioanl zeros of `f(x)` are:

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

We find:

`f(1/2) = 2/8+3/4+8/2-5 = 1/4+3/4+4-5 = 0`

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

`2x^3+3x^2+8x-5`

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

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

`= (2x-1)((x+1)^2-(2i)^2)`

`= (2x-1)(x+1-2i)(x+1+2i)`

Hence the other two zeros are:

`x = -1+-2i`