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

$3x^3+4x^2-7x+2=0$

The given equation is $3x^3+4x^2-7x+2=0$

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 $2$are

$p:\pm 1,\pm 2$

The factors of the leading coefficient $3$are

$q:\pm 1,\pm 3$

So, the possible rational zeros are

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

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

Thus,

$$\begin{gathered} f\left(1\right)=3{\left(1\right)}^3+4{\left(1\right)}^2-7\left(1\right)+2 \\ f\left(1\right)=3+4-7+2 \\ f\left(1\right)=2 \end{gathered}$$

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

Now, let's test the potential zero for $-1$by using substitution

Thus,

$$\begin{gathered} f(-1)=3{(-1)}^3+4{(-1)}^2-7(-1)+2 \\ f(-1)=-3+4+7+2 \\ f(-1)=10 \end{gathered}$$

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

Let's test the potential zero for $2$by using substitution

Thus,

$$\begin{gathered} f\left(2\right)=3{\left(2\right)}^3+4{\left(2\right)}^2-7\left(2\right)+2 \\ f\left(2\right)=24+16-14+2 \\ f\left(2\right)=28 \end{gathered}$$

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

Now, let's test the potential zero for $-2$by using substitution

Thus,

$$\begin{gathered} f(-2)=3{(-2)}^3+4{(-2)}^2-7(-2)+2 \\ f(-2)=-24+16+14+2 \\ f(-2)=8 \end{gathered}$$

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

Let's take the test for potential zero $\frac{1}{3}$by using substitution

Thus,

$$\begin{gathered} f\left(\frac{1}{3}\right)=3{\left(\frac{1}{3}\right)}^3+4{\left(\frac{1}{3}\right)}^2-7\left(\frac{1}{3}\right)+2 \\ f\left(\frac{1}{3}\right)=\frac{2}{9} \end{gathered}$$

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

Now, let's test for the potential zero $-\frac{1}{3}$by using substitution

Thus,

$$\begin{gathered} f\left(-\frac{1}{3}\right)=3{\left(-\frac{1}{3}\right)}^3+4{\left(-\frac{1}{3}\right)}^2-7\left(-\frac{1}{3}\right)+2 \\ f\left(\frac{1}{3}\right)=\frac{14}{3} \end{gathered}$$

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

Take the test of potential rational zero $\frac{2}{3}$by using synthetic division

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

The factor form is$\left(x-\frac{2}{3}\right)\left(3x^2+6x-3\right).$

As the remaining zeros satisfy the depressed equation

So,

$$\begin{gathered} 3x^2+6x-3=0 \\ 3(x^2+2x-1)=0 \\ \left(x+1-\sqrt{2}\right)\left(x+1+\sqrt{2}\right)=0 \\ x=-1+\sqrt{2},-1-\sqrt{2} \end{gathered}$$