Examples

f (x) = 9 x^{2} + 3 x - 3

$f (x) = 9 x^{2} + 3 x - 3$

Step 1

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

Largest Degree:

2

$2$

Leading Coefficient:

9

$9$

Step 2

Simplify each term.

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Step 2.1

Cancel the common factor of

9

$9$ .

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Step 2.1.1

Cancel the common factor.

f (x) = \frac{9 x^{2}}{9} + \frac{3 x}{9} + \frac{- 3}{9}

$f (x) = \frac{9 x^{2}}{9} + \frac{3 x}{9} + \frac{- 3}{9}$

Step 2.1.2

Divide

x^{2}

$x^{2}$ by

1

$1$ .

f (x) = x^{2} + \frac{3 x}{9} + \frac{- 3}{9}

$f (x) = x^{2} + \frac{3 x}{9} + \frac{- 3}{9}$

f (x) = x^{2} + \frac{3 x}{9} + \frac{- 3}{9}

$f (x) = x^{2} + \frac{3 x}{9} + \frac{- 3}{9}$

Step 2.2

Cancel the common factor of

3

$3$ and

9

$9$ .

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Step 2.2.1

Factor

3

$3$ out of

3 x

$3 x$ .

f (x) = x^{2} + \frac{3 (x)}{9} + \frac{- 3}{9}

$f (x) = x^{2} + \frac{3 (x)}{9} + \frac{- 3}{9}$

Step 2.2.2

Cancel the common factors.

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Step 2.2.2.1

Factor

3

$3$ out of

9

$9$ .

f (x) = x^{2} + \frac{3 x}{3 \cdot 3} + \frac{- 3}{9}

$f (x) = x^{2} + \frac{3 x}{3 \cdot 3} + \frac{- 3}{9}$

Step 2.2.2.2

Cancel the common factor.

f (x) = x^{2} + \frac{3 x}{3 \cdot 3} + \frac{- 3}{9}

$f (x) = x^{2} + \frac{3 x}{3 \cdot 3} + \frac{- 3}{9}$

Step 2.2.2.3

Rewrite the expression.

f (x) = x^{2} + \frac{x}{3} + \frac{- 3}{9}

$f (x) = x^{2} + \frac{x}{3} + \frac{- 3}{9}$

f (x) = x^{2} + \frac{x}{3} + \frac{- 3}{9}

$f (x) = x^{2} + \frac{x}{3} + \frac{- 3}{9}$

f (x) = x^{2} + \frac{x}{3} + \frac{- 3}{9}

$f (x) = x^{2} + \frac{x}{3} + \frac{- 3}{9}$

Step 2.3

Cancel the common factor of

- 3

$- 3$ and

9

$9$ .

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Step 2.3.1

Factor

3

$3$ out of

- 3

$- 3$ .

f (x) = x^{2} + \frac{x}{3} + \frac{3 (- 1)}{9}

$f (x) = x^{2} + \frac{x}{3} + \frac{3 (- 1)}{9}$

Step 2.3.2

Cancel the common factors.

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Step 2.3.2.1

Factor

3

$3$ out of

9

$9$ .

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

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

Step 2.3.2.2

Cancel the common factor.

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

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

Step 2.3.2.3

Rewrite the expression.

f (x) = x^{2} + \frac{x}{3} + \frac{- 1}{3}

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

f (x) = x^{2} + \frac{x}{3} + \frac{- 1}{3}

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

f (x) = x^{2} + \frac{x}{3} + \frac{- 1}{3}

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

Step 2.4

Move the negative in front of the fraction.

f (x) = x^{2} + \frac{x}{3} - \frac{1}{3}

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

Step 3

Create a list of the coefficients of the function except the leading coefficient of

$1$ .

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

Step 4

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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Step 4.1

Arrange the terms in ascending order.

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

Step 4.2

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

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

Step 4.3

$- \frac{1}{3}$ is approximately

$- 0.\overline{3}$ which is negative so negate

$- \frac{1}{3}$ and remove the absolute value

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

Step 4.4

Write

$1$ as a fraction with a common denominator.

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

Step 4.5

Combine the numerators over the common denominator.

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

Step 4.6

Add

$1$ and

$3$ .

$b_{1} = \frac{4}{3}$

Step 5

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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Step 5.1

Simplify each term.

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Step 5.1.1

$\frac{1}{3}$ is approximately

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

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

Step 5.1.2

$- \frac{1}{3}$ is approximately

$- 0.\overline{3}$ which is negative so negate

$- \frac{1}{3}$ and remove the absolute value

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

Step 5.2

Combine fractions.

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Step 5.2.1

Combine the numerators over the common denominator.

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

Step 5.2.2

Add

$1$ and

$1$ .

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

Step 5.3

Arrange the terms in ascending order.

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

Step 5.4

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

$b_{2} = 1$

Step 6

Take the smaller bound option between

$b_{1} = \frac{4}{3}$ and

$b_{2} = 1$ .

Smaller Bound:

$1$

Step 7

Every real root on

$f (x) = 9 x^{2} + 3 x - 3$ lies between

$- 1$ and

$1$ .

$- 1$ and

$1$

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