Precalculus Examples

$f (x) = x^{2} - 1$

Step 1

Find every combination of $\pm \frac{p}{q}$ .

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Step 1.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1$

$q = \pm 1$

Step 1.2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1$

Step 2

Apply synthetic division on $\frac{x^{2} - 1}{x - 1}$ when $x = 1$ .

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Step 2.1

Place the numbers representing the divisor and the dividend into a division-like configuration.

$1$	$1$	$0$	$- 1$

Step 2.2

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$1$	$1$	$0$	$- 1$

	$1$

Step 2.3

Multiply the newest entry in the result $(1)$ by the divisor $(1)$ and place the result of $(1)$ under the next term in the dividend $(0)$ .

$1$	$1$	$0$	$- 1$
		$1$
	$1$

Step 2.4

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$1$	$0$	$- 1$
		$1$
	$1$	$1$

Step 2.5

Multiply the newest entry in the result $(1)$ by the divisor $(1)$ and place the result of $(1)$ under the next term in the dividend $(- 1)$ .

$1$	$1$	$0$	$- 1$
		$1$	$1$
	$1$	$1$

Step 2.6

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$1$	$0$	$- 1$
		$1$	$1$
	$1$	$1$	$0$

Step 2.7

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$(1) x + 1$

Step 2.8

Simplify the quotient polynomial.

$x + 1$

Step 3

Since $1 > 0$ and all of the signs in the bottom row of the synthetic division are positive, $1$ is an upper bound for the real roots of the function.

Upper Bound: $1$

Step 4

Apply synthetic division on $\frac{x^{2} - 1}{x + 1}$ when $x = - 1$ .

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Step 4.1

Place the numbers representing the divisor and the dividend into a division-like configuration.

$- 1$	$1$	$0$	$- 1$

Step 4.2

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$- 1$	$1$	$0$	$- 1$

	$1$

Step 4.3

Multiply the newest entry in the result $(1)$ by the divisor $(- 1)$ and place the result of $(- 1)$ under the next term in the dividend $(0)$ .

$- 1$	$1$	$0$	$- 1$
		$- 1$
	$1$

Step 4.4

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 1$	$1$	$0$	$- 1$
		$- 1$
	$1$	$- 1$

Step 4.5

Multiply the newest entry in the result $(- 1)$ by the divisor $(- 1)$ and place the result of $(1)$ under the next term in the dividend $(- 1)$ .

$- 1$	$1$	$0$	$- 1$
		$- 1$	$1$
	$1$	$- 1$

Step 4.6

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 1$	$1$	$0$	$- 1$
		$- 1$	$1$
	$1$	$- 1$	$0$

Step 4.7

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$(1) x - 1$

Step 4.8

Simplify the quotient polynomial.

$x - 1$

Step 5

Since $- 1 < 0$ and the signs in the bottom row of the synthetic division alternate sign, $- 1$ is a lower bound for the real roots of the function.

Lower Bound: $- 1$

Step 6

Determine the upper and lower bounds.

Upper Bound: $1$

Lower Bound: $- 1$

Step 7

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