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Q. 65

Expert-verified
Found in: Page 210

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# In Problems 63–72, find the real solutions of each equation.$3{x}^{3}+4{x}^{2}-7x+2=0$

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

See the step by step solution

## Step 1. Finding the possible number of zeros

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

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

$p:±1,±2$

The factors of the leading coefficient $3$ are

$q:±1,±3$

So, the possible rational zeros are

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

## Step 3. Finding the zero

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

Thus,

$f\left(1\right)=3{\left(1\right)}^{3}+4{\left(1\right)}^{2}-7\left(1\right)+2\phantom{\rule{0ex}{0ex}}f\left(1\right)=3+4-7+2\phantom{\rule{0ex}{0ex}}f\left(1\right)=2$

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,

$f\left(-1\right)=3{\left(-1\right)}^{3}+4{\left(-1\right)}^{2}-7\left(-1\right)+2\phantom{\rule{0ex}{0ex}}f\left(-1\right)=-3+4+7+2\phantom{\rule{0ex}{0ex}}f\left(-1\right)=10$

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,

$f\left(2\right)=3{\left(2\right)}^{3}+4{\left(2\right)}^{2}-7\left(2\right)+2\phantom{\rule{0ex}{0ex}}f\left(2\right)=24+16-14+2\phantom{\rule{0ex}{0ex}}f\left(2\right)=28$

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,

$f\left(-2\right)=3{\left(-2\right)}^{3}+4{\left(-2\right)}^{2}-7\left(-2\right)+2\phantom{\rule{0ex}{0ex}}f\left(-2\right)=-24+16+14+2\phantom{\rule{0ex}{0ex}}f\left(-2\right)=8$

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,

$f\left(\frac{1}{3}\right)=3{\left(\frac{1}{3}\right)}^{3}+4{\left(\frac{1}{3}\right)}^{2}-7\left(\frac{1}{3}\right)+2\phantom{\rule{0ex}{0ex}}f\left(\frac{1}{3}\right)=\frac{2}{9}$

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,

$f\left(-\frac{1}{3}\right)=3{\left(-\frac{1}{3}\right)}^{3}+4{\left(-\frac{1}{3}\right)}^{2}-7\left(-\frac{1}{3}\right)+2\phantom{\rule{0ex}{0ex}}f\left(\frac{1}{3}\right)=\frac{14}{3}$

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

## Step 7. Use zero product property

As the remaining zeros satisfy the depressed equation

So,

$3{x}^{2}+6x-3=0\phantom{\rule{0ex}{0ex}}3\left({x}^{2}+2x-1\right)=0\phantom{\rule{0ex}{0ex}}\left(x+1-\sqrt{2}\right)\left(x+1+\sqrt{2}\right)=0\phantom{\rule{0ex}{0ex}}x=-1+\sqrt{2},-1-\sqrt{2}$