Question

How do you find all the zeros of `f(x)=3x^4-17x^3+33x^2-17x-10`?

Answer 1

Find zeros: `-1/3`, `2` and `2+-i`

Explanation:

By the rational root theorem, any rational zeros of `f(x)` must be expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-10` and `q` a divisor of the coefficient `3` of the leading term.

That means that the only possible rational zeros are:

`+-1/3, +-2/3, +-1, +-5/3, +-2, +-10/3, +-5, +-10`

Trying each in turn, we find:

`f(-1/3) = 3/81+17/27+33/9+17/3-10`

`=(1+17+99+153-270)/27 = 0`

`f(2) = 48-136+132-34+10 = 0`

So `x=-1/3` and `x=2` are zeros and `(3x+1)` and `(x-2)` are factors of `f(x)`:

`3x^4-17x^3+33x^2-17x-10`

`=(3x+1)(x^3-6x^2+13x-10)`

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

The remaining quadratic factor has negative discriminant, but we can factor it using Complex numbers, completeing the square and the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

as follows:

`x^2-4x+5 = (x-2)^2+1 = (x-2)^2-i^2 = ((x-2)-i)((x-2)+i) = (x-2-i)(x-2+i)`

So the remaining zeros are `x = 2+-i`