How do I discuss the zeros of a cubic equation in an open interval?

Question

Discuss the zeroes of $2x^3 - 3x^2 - 12x + 2 $ on the interval $ (- 2, 5). $

Comment: I am aware that this question is somewhat calculus related, but to go about it puzzles me.

Answer 1

In here you must apply the Bolzano's Theorem, that basically states that:

"If a continuous function defined on an interval is sometimes positive and sometimes negative, it must be 0 at some point."

Now let's see if there is a zero in $(-2,5)$:

$$f(-2)=-2$$ $$f(5)= 117$$

As we can see, do exist a root in this interval. Now we can apply some numerical methods in order to approximate (as we want to the) desired the root. Let's use the Bisection Method:

The steps are: 1st) we are going to divide the interval in which there is a zero; 2nd) evaluate $f$ in the middle point and see if it is positive or negative; 3rd) repeat again.

In this case the middle point is: $p_0=\dfrac{3}{2}$ Now let's follow the 2nd step: $f(1.5)=-16$ As we can see the new interval is $(1.5,5)$

$p_1=3.25$; $f(3.25)=-0.03125$; $(3.25,5)$.

$\cdots$ $\cdots$

As we can see, $3.25$ is already close to the root, however, only at infinity, you get exact result.

EDIT

A method that requires some calculus understanding is the Newton Method.