Pre-Algebra Examples

$p (x) = - x^{3} + 6 x^{2} - 4 x + 2$

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: $- 1$

Step 2

Simplify each term.

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Step 2.1

Dividing two negative values results in a positive value.

$p (x) = \frac{x^{3}}{1} + \frac{6 x^{2}}{- 1} + \frac{- 4 x}{- 1} + \frac{2}{- 1}$

Step 2.2

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

$p (x) = x^{3} + \frac{6 x^{2}}{- 1} + \frac{- 4 x}{- 1} + \frac{2}{- 1}$

Step 2.3

Move the negative one from the denominator of $\frac{6 x^{2}}{- 1}$ .

$p (x) = x^{3} - 1 \cdot (6 x^{2}) + \frac{- 4 x}{- 1} + \frac{2}{- 1}$

Step 2.4

Rewrite $- 1 \cdot (6 x^{2})$ as $- (6 x^{2})$ .

$p (x) = x^{3} - (6 x^{2}) + \frac{- 4 x}{- 1} + \frac{2}{- 1}$

Step 2.5

Multiply $6$ by $- 1$ .

$p (x) = x^{3} - 6 x^{2} + \frac{- 4 x}{- 1} + \frac{2}{- 1}$

Step 2.6

Move the negative one from the denominator of $\frac{- 4 x}{- 1}$ .

$p (x) = x^{3} - 6 x^{2} - 1 \cdot (- 4 x) + \frac{2}{- 1}$

Step 2.7

Rewrite $- 1 \cdot (- 4 x)$ as $- (- 4 x)$ .

$p (x) = x^{3} - 6 x^{2} - (- 4 x) + \frac{2}{- 1}$

Step 2.8

Multiply $- 4$ by $- 1$ .

$p (x) = x^{3} - 6 x^{2} + 4 x + \frac{2}{- 1}$

Step 2.9

Divide $2$ by $- 1$ .

$p (x) = x^{3} - 6 x^{2} + 4 x - 2$

Step 3

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

$- 6, 4, - 2$

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.

$b_{1} = | - 2 |, | 4 |, | - 6 |$

Step 4.2

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

$b_{1} = | - 6 |$

Step 4.3

The absolute value is the distance between a number and zero. The distance between $- 6$ and $0$ is $6$ .

$b_{1} = 6 + 1$

Step 4.4

Add $6$ and $1$ .

$b_{1} = 7$

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

The absolute value is the distance between a number and zero. The distance between $- 6$ and $0$ is $6$ .

$b_{2} = 6 + | 4 | + | - 2 |$

Step 5.1.2

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$b_{2} = 6 + 4 + | - 2 |$

Step 5.1.3

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

$b_{2} = 6 + 4 + 2$

Step 5.2

Simplify by adding numbers.

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Step 5.2.1

Add $6$ and $4$ .

$b_{2} = 10 + 2$

Step 5.2.2

Add $10$ and $2$ .

$b_{2} = 12$

Step 5.3

Arrange the terms in ascending order.

$b_{2} = 1, 12$

Step 5.4

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

$b_{2} = 12$

Step 6

Take the smaller bound option between $b_{1} = 7$ and $b_{2} = 12$ .

Smaller Bound: $7$

Step 7

Every real root on $p (x) = - x^{3} + 6 x^{2} - 4 x + 2$ lies between $- 7$ and $7$ .

$- 7$ and $7$