Question

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

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

Short answer

The real solution of the equation is$-\frac{3}{2}.$

Step-by-step solution

Step 1. Finding the possible number of zeros 

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.

Step 2. Use the rational zero theorem 

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

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

Step 3. Finding the zero 

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

Step 4. Finding the other zero 

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.

Step 5. Use synthetic division

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 $-\frac{3}{2}.$