How do you use the intermediate value theorem to show that there is a root of the equation $2{x^3} + {x^2} + 2 = 0$ over the interval $\left( { - 2, - 1} \right)$?

Hint: In this question we will first find the values of the function on the given intervals and check whether they fulfill the requirements of the intermediate value theorem, if they do, we can conclude that there is a root which exists in the interval given in the question.

Complete step-by-step solution:
We have the given equation:
$2{x^3} + {x^2} + 2 = 0$
Let’s consider the equation to be a function of $x$ therefore, we can write it as:
$ \Rightarrow f(x) = 2{x^3} + {x^2} + 2$
Now the intermediate value theorem says that for a function $f(x)$ on the interval $[a,b]$if the value of $f(a) < 0$ and the value of $f(b) > 0$ or the value of $f(a) > 0$ and the value of $f(b) < 0$, then there exists a point $c$ which has value which is in between the two points $a$ and $b$ such that $f(c) = 0$
Now the value of $f( - 2)$ can be found out as:
$ \Rightarrow f( - 2) = 2{( - 2)^3} + {( - 2)^2} + 2$
On simplifying we get:
$ \Rightarrow f( - 2) = 2( - 8) + 4 + 2$
On adding all the terms, we get:
$ \Rightarrow f( - 2) = - 10$
Now the value of $f( - 1)$ can be found out as:
$ \Rightarrow f( - 1) = 2{( - 1)^3} + {( - 1)^2} + 2$
On simplifying we get:
$ \Rightarrow f( - 2) = 2( - 1) + 1 + 2$
On adding all the terms, we get:
$ \Rightarrow f( - 2) = 1$

Now because the value of $f( - 2) < 0$ and the value of $f( - 1) > 0$ and the function $f(x)$ is continuous on the closed interval $[ - 2, - 1]$, the expression satisfies the conditions of the intermediate value theorem therefore, there exists a point $c$ on the interval $[ - 2, - 1]$ for which $f(c) = 0$.

Note: It is to be noted that $f(x)$ is a continuous function since it is a polynomial cubic function therefore; it is continuous at each point in the interval.
It is to be remembered that the square of a negative number yields a positive number because the multiplication of two negative numbers is positive.