Question

How do you find all zeros of `F(x)= x^3 -8x^2 +25x -26`?

Answer 1

There are three solutions:
`x_0 = 2`
`x_1 = 3+2i`
`x_2 = 3-2i`

Explanation:

The rational root theorem tells us that rational roots to a polynomial equation with integer coefficients can be written in the form `p/q`, where `p` is a factor of the constant term and `q` is a factor of the leading coefficient.

The polynomial equation is `1*x^3 - 8x^2 + 25x - 26 = 0`.
The integer factors of the constant `-26` are `+-26, +-13,+-2,+-1`.
The integer factors of the leading coefficient `1` are `+-1`.

The rational solutions must therefore be among the following: `+-26/1, +-13/1, +-2/1, +-1/1`

Try each answer by substituting them into `x` in the equation until you find the solution. You will find that `x = 2` solves the equation. This is the first solution.

Now we know that `(x-2)` is a factor of the polynomial, meaning that `x^3 - 8x^2 + 25x - 26 = (x-2)*g(x)` where `g(x)` is some polynomial of a lower degree. Use polynomial division to find:

`g(x) = (x^3 - 8x^2 + 25x - 26) / (x-2)=x^2−6x+13`

(Polynomial divison might need its own explanation for those who haven't learned it. If there's a separate Socratic thread on it, it should be linked here.)

With the factorized polynomial `(x-2)(x^2−6x+13) = 0`, now solve for `x` when `(x^2−6x+13) = 0`

A quadratic equation `ax^2+bx+c=0` can be solved using the quadratic formula: `x = (-b+- sqrt(b^2 -4ac))/(2a)`.

Plug `a = 1`, `b =-6` and `c = 13` into it and we get:

`x = 3+-sqrt(-4)`
`= 3+-sqrt(4*-1)`
`= 3+-sqrt(4)sqrt(-1)`
`= 3+-2i`

Now we have found all three solutions:
`x_0 = 2`
`x_1 = 3+2i`
`x_2 = 3-2i`