Find the Bounds of the Zeros f(x)=-15x^3+30x^2+17x-45

$f (x) = - 15 x^{3} + 30 x^{2} + 17 x - 45$

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

Largest Degree: $3$

Leading Coefficient: $- 15$

Simplify each term.

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Cancel the common factor of $- 15$ .

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Cancel the common factor.

$f (x) = \frac{- 15 x^{3}}{- 15} + \frac{30 x^{2}}{- 15} + \frac{17 x}{- 15} + \frac{- 45}{- 15}$

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

$f (x) = x^{3} + \frac{30 x^{2}}{- 15} + \frac{17 x}{- 15} + \frac{- 45}{- 15}$

Cancel the common factor of $30$ and $- 15$ .

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Factor $15$ out of $30 x^{2}$ .

$f (x) = x^{3} + \frac{15 (2 x^{2})}{- 15} + \frac{17 x}{- 15} + \frac{- 45}{- 15}$

Move the negative one from the denominator of $\frac{2 x^{2}}{- 1}$ .

$f (x) = x^{3} - 1 \cdot (2 x^{2}) + \frac{17 x}{- 15} + \frac{- 45}{- 15}$

Rewrite $- 1 \cdot (2 x^{2})$ as $- (2 x^{2})$ .

$f (x) = x^{3} - (2 x^{2}) + \frac{17 x}{- 15} + \frac{- 45}{- 15}$

Multiply $2$ by $- 1$ .

$f (x) = x^{3} - 2 x^{2} + \frac{17 x}{- 15} + \frac{- 45}{- 15}$

Move the negative in front of the fraction.

$f (x) = x^{3} - 2 x^{2} - \frac{17 x}{15} + \frac{- 45}{- 15}$

Divide $- 45$ by $- 15$ .

$f (x) = x^{3} - 2 x^{2} - \frac{17 x}{15} + 3$

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

$- 2, - \frac{17}{15}, 3$

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.

$b_{1} = | - \frac{17}{15} |, | - 2 |, | 3 |$

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

$b_{1} = | 3 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$b_{1} = 3 + 1$

Add $3$ and $1$ .

$b_{1} = 4$

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 each term.

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The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$b_{2} = 2 + | - \frac{17}{15} | + | 3 |$

$- \frac{17}{15}$ is approximately $- 1.1\overline{3}$ which is negative so negate $- \frac{17}{15}$ and remove the absolute value

$b_{2} = 2 + \frac{17}{15} + | 3 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$b_{2} = 2 + \frac{17}{15} + 3$

Find the common denominator.

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Write $2$ as a fraction with denominator $1$ .

$b_{2} = \frac{2}{1} + \frac{17}{15} + 3$

Multiply $\frac{2}{1}$ by $\frac{15}{15}$ .

$b_{2} = \frac{2}{1} \cdot \frac{15}{15} + \frac{17}{15} + 3$

Multiply $\frac{2}{1}$ and $\frac{15}{15}$ .

$b_{2} = \frac{2 \cdot 15}{15} + \frac{17}{15} + 3$

Write $3$ as a fraction with denominator $1$ .

$b_{2} = \frac{2 \cdot 15}{15} + \frac{17}{15} + \frac{3}{1}$

Multiply $\frac{3}{1}$ by $\frac{15}{15}$ .

$b_{2} = \frac{2 \cdot 15}{15} + \frac{17}{15} + \frac{3}{1} \cdot \frac{15}{15}$

Multiply $\frac{3}{1}$ and $\frac{15}{15}$ .

$b_{2} = \frac{2 \cdot 15}{15} + \frac{17}{15} + \frac{3 \cdot 15}{15}$

Combine fractions.

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Combine fractions with similar denominators.

$b_{2} = \frac{2 \cdot 15 + 17 + 3 \cdot 15}{15}$

Multiply.

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Multiply $2$ by $15$ .

$b_{2} = \frac{30 + 17 + 3 \cdot 15}{15}$

Multiply $3$ by $15$ .

$b_{2} = \frac{30 + 17 + 45}{15}$

Simplify the numerator.

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Add $30$ and $17$ .

$b_{2} = \frac{47 + 45}{15}$

Add $47$ and $45$ .

$b_{2} = \frac{92}{15}$

Arrange the terms in ascending order.

$b_{2} = 1, \frac{92}{15}$

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

$b_{2} = \frac{92}{15}$

Take the smaller bound option between $b_{1} = 4$ and $b_{2} = \frac{92}{15}$ .

Smaller Bound: $4$

Every real root on $f (x) = - 15 x^{3} + 30 x^{2} + 17 x - 45$ lies between $- 4$ and $4$ .

$- 4$ and $4$

Find the Bounds of the Zeros f(x)=-15x^3+30x^2+17x-45

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