Question

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

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

Short answer

The real solution of the equation is$\frac{5}{2}.$

Step-by-step solution

Step 1. Finding the possible number of zeros 

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.

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 $-5$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

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

Step 3. Use the synthetic division 

Let's test the potential zero $1$,

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

Step 4. Finding the zero 

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$\left(2x-5\right)\left(2x^2+2x+2\right).$

Step 5. Use the zero product property 

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$\frac{5}{2}.$