Pre-Algebra Examples

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

Step 1

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

Largest Degree: $2$

Leading Coefficient: $3$

Step 2

Simplify each term.

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Step 2.1

Cancel the common factor of $3$ .

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Step 2.1.1

Cancel the common factor.

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

Step 2.1.2

Divide $x^{2}$ by $1$ .

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

Step 2.2

Move the negative in front of the fraction.

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

Step 3

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

$- \frac{2}{3}$

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.

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

Step 4.2

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

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

Step 4.3

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

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

Step 4.4

Combine the numerators over the common denominator.

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

Step 4.5

Add $2$ and $3$ .

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

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

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

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

Step 5.2

Arrange the terms in ascending order.

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

Step 5.3

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

$b_{2} = 1$

Step 6

Take the smaller bound option between $b_{1} = \frac{5}{3}$ and $b_{2} = 1$ .

Smaller Bound: $1$

Step 7

Every real root on $f (x) = 3 x^{2} - 2 x$ lies between $- 1$ and $1$ .

$- 1$ and $1$