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

$2x^3+3x^2+2x+3=0$

The given equation is $2x^3+3x^2+2x+3=0$

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 $3$are

$p:\pm 1,\pm 3$

The factors of the leading coefficient $2$are

$q:\hspace{0.17em}\pm 1,\pm 2$

So, the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2}$

Let's test the potential zero $1$by using substitution

Thus,

$$\begin{gathered} f\left(1\right)=2{\left(1\right)}^3+3{\left(1\right)}^2+2\left(1\right)+3 \\ f\left(1\right)=2+3+2+3 \\ f\left(1\right)+10 \end{gathered}$$

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

Now, let's test the potential zero for $-\frac{3}{2}$by using substitution

Thus,

$$\begin{gathered} f\left(-\frac{3}{2}\right)=2{\left(-\frac{3}{2}\right)}^3+3{\left(-\frac{3}{2}\right)}^2+2\left(-\frac{3}{2}\right)+3 \\ f\left(-\frac{3}{2}\right)=0 \end{gathered}$$

So, $-\frac{3}{2}$is zero and $\left(x+\frac{3}{2}\right)or(2x+3)$is a factor.

Let's factor the equation

By factoring the depressed equation we get is $2x^2+2=0.$

As we know the other zeros satisfy the depressed equation and the depressed equation is a quadratic equation with discriminant $b^2-4ac=0-16$which is less than zero. Thus, the equation has no real solution other than $-\frac{3}{2}.$