Pre-Algebra Examples

$f (x) = \frac{2}{3} x - 12$

Step 1

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

Largest Degree: $1$

Leading Coefficient: $1$

Step 2

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

$\frac{2}{3}, - 12$

Step 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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Step 3.1

Arrange the terms in ascending order.

$b_{1} = | \frac{2}{3} |, | - 12 |$

Step 3.2

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

$b_{1} = | - 12 |$

Step 3.3

The absolute value is the distance between a number and zero. The distance between $- 12$ and $0$ is $12$ .

$b_{1} = 12 + 1$

Step 3.4

Add $12$ and $1$ .

$b_{1} = 13$

Step 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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Step 4.1

Simplify each term.

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Step 4.1.1

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

$b_{2} = \frac{2}{3} + | - 12 |$

Step 4.1.2

The absolute value is the distance between a number and zero. The distance between $- 12$ and $0$ is $12$ .

$b_{2} = \frac{2}{3} + 12$

Step 4.2

To write $12$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$b_{2} = \frac{2}{3} + 12 \cdot \frac{3}{3}$

Step 4.3

Combine $12$ and $\frac{3}{3}$ .

$b_{2} = \frac{2}{3} + \frac{12 \cdot 3}{3}$

Step 4.4

Combine the numerators over the common denominator.

$b_{2} = \frac{2 + 12 \cdot 3}{3}$

Step 4.5

Simplify the numerator.

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Step 4.5.1

Multiply $12$ by $3$ .

$b_{2} = \frac{2 + 36}{3}$

Step 4.5.2

Add $2$ and $36$ .

$b_{2} = \frac{38}{3}$

Step 4.6

Arrange the terms in ascending order.

$b_{2} = 1, \frac{38}{3}$

Step 4.7

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

$b_{2} = \frac{38}{3}$

Step 5

Take the smaller bound option between $b_{1} = 13$ and $b_{2} = \frac{38}{3}$ .

Smaller Bound: $\frac{38}{3}$

Step 6

Every real root on $f (x) = \frac{2}{3} x - 12$ lies between $- \frac{38}{3}$ and $\frac{38}{3}$ .

$- \frac{38}{3}$ and $\frac{38}{3}$