Question

How do you find all zeros with multiplicities of `f(x)=3x^3+3x^2-11x-10`?

Answer 1

The zeros are `-2`, `1/2+sqrt(69)/6`, `1/2-sqrt(69)/6` all with multiplicity `1`.

Explanation:

Given:

`f(x) = 3x^3+3x^2-11x-10`

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 `-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 eventually find:

`f(-2) = 3(color(blue)(-2))^3+3(color(blue)(-2))^2-11(color(blue)(-2))-10`

`color(white)(f(-2)) = -24+12+22-10`

`color(white)(f(-2)) = 0`

So `-2` is a zero and `(x+2)` a factor:

`3x^3+3x^2-11x-10 = (x+2)(3x^2-3x-5)`

The remaining quadratic factor is in the form:

`ax^2+bx+c`

with `a=3`, `b=-3` and `c=-5`

We can find its zeros using the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x) = (3+-sqrt((-3)^2-4(3)(-5)))/(2*3)`

`color(white)(x) = (3+-sqrt(9+60))/(2*3)`

`color(white)(x) = 1/2+-sqrt(69)/6`