Question

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

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

Short answer

The real solution of the equation are$\frac{2}{3},(-1+\sqrt{2}),and(-1-\sqrt{2}).$

Step-by-step solution

Step 1. Finding the possible number of zeros 

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.

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 $2$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

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

Step 3. Finding the zero  

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.

Step 4. Finding the zero 

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.

Step 5. Finding the zero 

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.

Step 6. Finding the zero

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

Step 7. Use zero product property

As the remaining zeros satisfy the depressed equation

So,

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