Find the Bounds of the Zeros f(x)=3x^3-2x^2+x+2

$f (x) = 3 x^{3} - 2 x^{2} + x + 2$

Check the leading coefficient of the function. This number is the coefficient of the expression with the largest degree.

Largest Degree: $3$

Leading Coefficient: $3$

Simplify each term.

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Cancel the common factor of $3$ .

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Cancel the common factor.

$f (x) = \frac{3 x^{3}}{3} + \frac{- 2 x^{2}}{3} + \frac{x}{3} + \frac{2}{3}$

Divide $x^{3}$ by $1$ .

$f (x) = x^{3} + \frac{- 2 x^{2}}{3} + \frac{x}{3} + \frac{2}{3}$

Move the negative in front of the fraction.

$f (x) = x^{3} - \frac{2 x^{2}}{3} + \frac{x}{3} + \frac{2}{3}$

Create a list of the coefficients of the function except the leading coefficient of $1$ .

$- \frac{2}{3}, \frac{1}{3}, \frac{2}{3}$

There will be two bound options, $b_{1}$ and $b_{2}$ , the smaller of which is the answer. To calculate the first bound option, find the absolute value of the largest coefficient from the list of coefficients. Then add $1$ .

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Arrange the terms in ascending order.

$b_{1} = | \frac{1}{3} |, | - \frac{2}{3} |, | \frac{2}{3} |$

The maximum value is the largest value in the arranged data set.

$b_{1} = | \frac{2}{3} |$

$\frac{2}{3}$ is approximately $0.\overline{6}$ which is positive so remove the absolute value

$b_{1} = \frac{2}{3} + 1$

Write $1$ as a fraction with a common denominator.

$b_{1} = \frac{2}{3} + \frac{3}{3}$

Combine the numerators over the common denominator.

$b_{1} = \frac{2 + 3}{3}$

Add $2$ and $3$ .

$b_{1} = \frac{5}{3}$

To calculate the second bound option, sum the absolute values of the coefficients from the list of coefficients. If the sum is greater than $1$ , use that number. If not, use $1$ .

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Simplify terms.

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Simplify each term.

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$- \frac{2}{3}$ is approximately $- 0.\overline{6}$ which is negative so negate $- \frac{2}{3}$ and remove the absolute value

$b_{2} = \frac{2}{3} + | \frac{1}{3} | + | \frac{2}{3} |$

$\frac{1}{3}$ is approximately $0.\overline{3}$ which is positive so remove the absolute value

$b_{2} = \frac{2}{3} + \frac{1}{3} + | \frac{2}{3} |$

$\frac{2}{3}$ is approximately $0.\overline{6}$ which is positive so remove the absolute value

$b_{2} = \frac{2}{3} + \frac{1}{3} + \frac{2}{3}$

Combine fractions with similar denominators.

$b_{2} = \frac{2 + 1 + 2}{3}$

Simplify the numerator.

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Add $2$ and $1$ .

$b_{2} = \frac{3 + 2}{3}$

Add $3$ and $2$ .

$b_{2} = \frac{5}{3}$

Arrange the terms in ascending order.

$b_{2} = 1, \frac{5}{3}$

The maximum value is the largest value in the arranged data set.

$b_{2} = \frac{5}{3}$

The bound options are the same.

Bound: $\frac{5}{3}$

Every real root on $f (x) = 3 x^{3} - 2 x^{2} + x + 2$ lies between $- \frac{5}{3}$ and $\frac{5}{3}$ .

$- \frac{5}{3}$ and $\frac{5}{3}$

Find the Bounds of the Zeros f(x)=3x^3-2x^2+x+2

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