Pre-Algebra Examples

$f (x) = 6 x^{3} - x^{2} - 294 x + 49$

Step 1

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

Largest Degree: $3$

Leading Coefficient: $6$

Step 2

Simplify each term.

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Step 2.1

Cancel the common factor of $6$ .

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Step 2.1.1

Cancel the common factor.

$f (x) = \frac{6 x^{3}}{6} + \frac{- x^{2}}{6} + \frac{- 294 x}{6} + \frac{49}{6}$

Step 2.1.2

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

$f (x) = x^{3} + \frac{- x^{2}}{6} + \frac{- 294 x}{6} + \frac{49}{6}$

Step 2.2

Move the negative in front of the fraction.

$f (x) = x^{3} - \frac{x^{2}}{6} + \frac{- 294 x}{6} + \frac{49}{6}$

Step 2.3

Cancel the common factor of $- 294$ and $6$ .

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Step 2.3.1

Factor $6$ out of $- 294 x$ .

$f (x) = x^{3} - \frac{x^{2}}{6} + \frac{6 (- 49 x)}{6} + \frac{49}{6}$

Step 2.3.2

Cancel the common factors.

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Step 2.3.2.1

Factor $6$ out of $6$ .

$f (x) = x^{3} - \frac{x^{2}}{6} + \frac{6 (- 49 x)}{6 (1)} + \frac{49}{6}$

Step 2.3.2.2

Cancel the common factor.

$f (x) = x^{3} - \frac{x^{2}}{6} + \frac{6 (- 49 x)}{6 \cdot 1} + \frac{49}{6}$

Step 2.3.2.3

Rewrite the expression.

$f (x) = x^{3} - \frac{x^{2}}{6} + \frac{- 49 x}{1} + \frac{49}{6}$

Step 2.3.2.4

Divide $- 49 x$ by $1$ .

$f (x) = x^{3} - \frac{x^{2}}{6} - 49 x + \frac{49}{6}$

Step 3

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

$- \frac{1}{6}, - 49, \frac{49}{6}$

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.

$b_{1} = | - \frac{1}{6} |, | \frac{49}{6} |, | - 49 |$

Step 4.2

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

$b_{1} = | - 49 |$

Step 4.3

The absolute value is the distance between a number and zero. The distance between $- 49$ and $0$ is $49$ .

$b_{1} = 49 + 1$

Step 4.4

Add $49$ and $1$ .

$b_{1} = 50$

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

$- \frac{1}{6}$ is approximately $- 0.1\overline{6}$ which is negative so negate $- \frac{1}{6}$ and remove the absolute value

$b_{2} = \frac{1}{6} + | - 49 | + | \frac{49}{6} |$

Step 5.1.2

The absolute value is the distance between a number and zero. The distance between $- 49$ and $0$ is $49$ .

$b_{2} = \frac{1}{6} + 49 + | \frac{49}{6} |$

Step 5.1.3

$\frac{49}{6}$ is approximately $8.1\overline{6}$ which is positive so remove the absolute value

$b_{2} = \frac{1}{6} + 49 + \frac{49}{6}$

Step 5.2

Simplify terms.

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Step 5.2.1

Combine the numerators over the common denominator.

$b_{2} = 49 + \frac{1 + 49}{6}$

Step 5.2.2

Add $1$ and $49$ .

$b_{2} = 49 + \frac{50}{6}$

Step 5.2.3

Cancel the common factor of $50$ and $6$ .

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Step 5.2.3.1

Factor $2$ out of $50$ .

$b_{2} = 49 + \frac{2 (25)}{6}$

Step 5.2.3.2

Cancel the common factors.

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Step 5.2.3.2.1

Factor $2$ out of $6$ .

$b_{2} = 49 + \frac{2 \cdot 25}{2 \cdot 3}$

Step 5.2.3.2.2

Cancel the common factor.

$b_{2} = 49 + \frac{2 \cdot 25}{2 \cdot 3}$

Step 5.2.3.2.3

Rewrite the expression.

$b_{2} = 49 + \frac{25}{3}$

Step 5.3

To write $49$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$b_{2} = 49 \cdot \frac{3}{3} + \frac{25}{3}$

Step 5.4

Combine $49$ and $\frac{3}{3}$ .

$b_{2} = \frac{49 \cdot 3}{3} + \frac{25}{3}$

Step 5.5

Combine the numerators over the common denominator.

$b_{2} = \frac{49 \cdot 3 + 25}{3}$

Step 5.6

Simplify the numerator.

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Step 5.6.1

Multiply $49$ by $3$ .

$b_{2} = \frac{147 + 25}{3}$

Step 5.6.2

Add $147$ and $25$ .

$b_{2} = \frac{172}{3}$

Step 5.7

Arrange the terms in ascending order.

$b_{2} = 1, \frac{172}{3}$

Step 5.8

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

$b_{2} = \frac{172}{3}$

Step 6

Take the smaller bound option between $b_{1} = 50$ and $b_{2} = \frac{172}{3}$ .

Smaller Bound: $50$

Step 7

Every real root on $f (x) = 6 x^{3} - x^{2} - 294 x + 49$ lies between $- 50$ and $50$ .

$- 50$ and $50$