In Problems 63–72, find the real solutions of each equation.

$2x^3-3x^2-3x-5=0$

The given equation is $2x^3-3x^2-3x-5=0$

The given function is of degree three, so it has at most three real zeros.

As all the coefficients are integers so we use the rational zeros theorem.

The factors of the constant term $-5$are

$p:\pm 1,\pm 5$

The factors of the leading coefficient $2$are

$q:\pm 1,\pm 2$

So, the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 5,\pm \frac{1}{2},\pm \frac{5}{2}$

Let's test the potential zero $1$,

Since the remainder is not zero, it is not a factor.

Now, let's test the potential zero $\frac{5}{2}$ by using the synthetic division test

Since the remainder is zero, $\left(x-\frac{5}{2}\right)=(2x-5)$is a factor.

The factor form is$\left(2x-5\right)\left(2x^2+2x+2\right).$

As the remaining zeros satisfy the depressed equation and the depressed equation is a quadratic equation with discriminant

$$\begin{gathered} b^2-4ac \\ =4-4\left(2\right)\left(2\right) \\ =-12 \end{gathered}$$which is less than zero. Thus, the equation has no real solutions other than$\frac{5}{2}.$