Precalculus Examples

$x^{4} - 1$

Step 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 2

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

$\pm 1$

Step 3

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

${(1)}^{4} - 1$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 1$ is a root of the polynomial.

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Step 4.1

One to any power is one.

$1 - 1$

Step 4.2

Subtract $1$ from $1$ .

$0$

Step 5

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{4} - 1}{x - 1}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

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Step 6.1

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

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

Step 6.2

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

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

	$1$

Step 6.3

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 $(0)$ .

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

Step 6.4

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

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

Step 6.5

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 $(0)$ .

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

Step 6.6

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

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

Step 6.7

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 $(0)$ .

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

Step 6.8

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

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

Step 6.9

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 $(- 1)$ .

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

Step 6.10

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

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

Step 6.11

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

$1 x^{3} + 1 x^{2} + (1) x + 1$

Step 6.12

Simplify the quotient polynomial.

$x^{3} + x^{2} + x + 1$

Step 7

Factor out the greatest common factor from each group.

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Step 7.1

Group the first two terms and the last two terms.

$(x - 1) (x^{3} + x^{2}) + x + 1$

Step 7.2

Factor out the greatest common factor (GCF) from each group.

$(x - 1) + x^{2} (x + 1) + 1 (x + 1)$

$(x - 1) x^{2} (x + 1) + 1 (x + 1)$

Step 8

Factor the polynomial by factoring out the greatest common factor, $x + 1$ .

$(x - 1) (x + 1) (x^{2} + 1)$

Step 9

Add $1$ to both sides of the equation.

$x^{4} = 1$

Step 10

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt[4]{1}$

Step 11

Any root of $1$ is $1$ .

$x = \pm 1$

Step 12

The complete solution is the result of both the positive and negative portions of the solution.

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Step 12.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Step 12.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

Step 12.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

Step 13