Precalculus Examples

$x^{4} - 2 x^{3} + x^{2} - 8 x - 12$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

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.

${(- 1)}^{4} - 2 {(- 1)}^{3} + {(- 1)}^{2} - 8 \cdot - 1 - 12$

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

Simplify each term.

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Step 4.1.1

Raise $- 1$ to the power of $4$ .

$1 - 2 {(- 1)}^{3} + {(- 1)}^{2} - 8 \cdot - 1 - 12$

Step 4.1.2

Raise $- 1$ to the power of $3$ .

$1 - 2 \cdot - 1 + {(- 1)}^{2} - 8 \cdot - 1 - 12$

Step 4.1.3

Multiply $- 2$ by $- 1$ .

$1 + 2 + {(- 1)}^{2} - 8 \cdot - 1 - 12$

Step 4.1.4

Raise $- 1$ to the power of $2$ .

$1 + 2 + 1 - 8 \cdot - 1 - 12$

Step 4.1.5

Multiply $- 8$ by $- 1$ .

$1 + 2 + 1 + 8 - 12$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Add $1$ and $2$ .

$3 + 1 + 8 - 12$

Step 4.2.2

Add $3$ and $1$ .

$4 + 8 - 12$

Step 4.2.3

Add $4$ and $8$ .

$12 - 12$

Step 4.2.4

Subtract $12$ from $12$ .

$0$

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.

$\frac{x^{4} - 2 x^{3} + x^{2} - 8 x - 12}{x + 1}$

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.

$- 1$	$1$	$- 2$	$1$	$- 8$	$- 12$

Step 6.2

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

$- 1$	$1$	$- 2$	$1$	$- 8$	$- 12$

	$1$

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 $(- 2)$ .

$- 1$	$1$	$- 2$	$1$	$- 8$	$- 12$
		$- 1$
	$1$

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.

$- 1$	$1$	$- 2$	$1$	$- 8$	$- 12$
		$- 1$
	$1$	$- 3$

Step 6.5

Multiply the newest entry in the result $(- 3)$ by the divisor $(- 1)$ and place the result of $(3)$ under the next term in the dividend $(1)$ .

$- 1$	$1$	$- 2$	$1$	$- 8$	$- 12$
		$- 1$	$3$
	$1$	$- 3$

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.

$- 1$	$1$	$- 2$	$1$	$- 8$	$- 12$
		$- 1$	$3$
	$1$	$- 3$	$4$

Step 6.7

Multiply the newest entry in the result $(4)$ by the divisor $(- 1)$ and place the result of $(- 4)$ under the next term in the dividend $(- 8)$ .

$- 1$	$1$	$- 2$	$1$	$- 8$	$- 12$
		$- 1$	$3$	$- 4$
	$1$	$- 3$	$4$

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.

$- 1$	$1$	$- 2$	$1$	$- 8$	$- 12$
		$- 1$	$3$	$- 4$
	$1$	$- 3$	$4$	$- 12$

Step 6.9

Multiply the newest entry in the result $(- 12)$ by the divisor $(- 1)$ and place the result of $(12)$ under the next term in the dividend $(- 12)$ .

$- 1$	$1$	$- 2$	$1$	$- 8$	$- 12$
		$- 1$	$3$	$- 4$	$12$
	$1$	$- 3$	$4$	$- 12$

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.

$- 1$	$1$	$- 2$	$1$	$- 8$	$- 12$
		$- 1$	$3$	$- 4$	$12$
	$1$	$- 3$	$4$	$- 12$	$0$

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.

$1 x^{3} + - 3 x^{2} + (4) x - 12$

Step 6.12

Simplify the quotient polynomial.

$x^{3} - 3 x^{2} + 4 x - 12$

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.

$(x + 1) (x^{3} - 3 x^{2}) + 4 x - 12$

Step 7.2

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

$(x + 1) + x^{2} (x - 3) + 4 (x - 3)$

$(x + 1) x^{2} (x - 3) + 4 (x - 3)$

Step 8

Factor the polynomial by factoring out the greatest common factor, $x - 3$ .

$(x + 1) (x - 3) (x^{2} + 4)$

Step 9

Factor the left side of the equation.

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Step 9.1

Regroup terms.

$- 2 x^{3} - 8 x + x^{4} + x^{2} - 12 = 0$

Step 9.2

Factor $- 2 x$ out of $- 2 x^{3} - 8 x$ .

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Step 9.2.1

Factor $- 2 x$ out of $- 2 x^{3}$ .

$- 2 x \cdot x^{2} - 8 x + x^{4} + x^{2} - 12 = 0$

Step 9.2.2

Factor $- 2 x$ out of $- 8 x$ .

$- 2 x \cdot x^{2} - 2 x \cdot 4 + x^{4} + x^{2} - 12 = 0$

Step 9.2.3

Factor $- 2 x$ out of $- 2 x (x^{2}) - 2 x (4)$ .

$- 2 x (x^{2} + 4) + x^{4} + x^{2} - 12 = 0$

Step 9.3

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$- 2 x (x^{2} + 4) + {(x^{2})}^{2} + x^{2} - 12 = 0$

Step 9.4

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$- 2 x (x^{2} + 4) + u^{2} + u - 12 = 0$

Step 9.5

Factor $u^{2} + u - 12$ using the AC method.

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Step 9.5.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $1$ .

$- 3, 4$

Step 9.5.2

Write the factored form using these integers.

$- 2 x (x^{2} + 4) + (u - 3) (u + 4) = 0$

Step 9.6

Replace all occurrences of $u$ with $x^{2}$ .

$- 2 x (x^{2} + 4) + (x^{2} - 3) (x^{2} + 4) = 0$

Step 9.7

Factor $x^{2} + 4$ out of $- 2 x (x^{2} + 4) + (x^{2} - 3) (x^{2} + 4)$ .

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Step 9.7.1

Factor $x^{2} + 4$ out of $- 2 x (x^{2} + 4)$ .

$(x^{2} + 4) (- 2 x) + (x^{2} - 3) (x^{2} + 4) = 0$

Step 9.7.2

Factor $x^{2} + 4$ out of $(x^{2} - 3) (x^{2} + 4)$ .

$(x^{2} + 4) (- 2 x) + (x^{2} + 4) (x^{2} - 3) = 0$

Step 9.7.3

Factor $x^{2} + 4$ out of $(x^{2} + 4) (- 2 x) + (x^{2} + 4) (x^{2} - 3)$ .

$(x^{2} + 4) (- 2 x + x^{2} - 3) = 0$

Step 9.8

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$(x^{2} + 4) (- 2 u + u^{2} - 3) = 0$

Step 9.9

Factor $- 2 u + u^{2} - 3$ using the AC method.

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Step 9.9.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 9.9.2

Write the factored form using these integers.

$(x^{2} + 4) ((u - 3) (u + 1)) = 0$

Step 9.10

Factor.

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Step 9.10.1

Replace all occurrences of $u$ with $x$ .

$(x^{2} + 4) ((x - 3) (x + 1)) = 0$

Step 9.10.2

Remove unnecessary parentheses.

$(x^{2} + 4) (x - 3) (x + 1) = 0$

Step 10

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x^{2} + 4 = 0$

$x - 3 = 0$

$x + 1 = 0$

Step 11

Set $x^{2} + 4$ equal to $0$ and solve for $x$ .

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Step 11.1

Set $x^{2} + 4$ equal to $0$ .

$x^{2} + 4 = 0$

Step 11.2

Solve $x^{2} + 4 = 0$ for $x$ .

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Step 11.2.1

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

Step 11.2.2

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

$x = \pm \sqrt{- 4}$

Step 11.2.3

Simplify $\pm \sqrt{- 4}$ .

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Step 11.2.3.1

Rewrite $- 4$ as $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Step 11.2.3.2

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{4}$

Step 11.2.3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{4}$

Step 11.2.3.4

Rewrite $4$ as $2^{2}$ .

$x = \pm i \cdot \sqrt{2^{2}}$

Step 11.2.3.5

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 2$

Step 11.2.3.6

Move $2$ to the left of $i$ .

$x = \pm 2 i$

Step 11.2.4

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

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Step 11.2.4.1

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

$x = 2 i$

Step 11.2.4.2

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

$x = - 2 i$

Step 11.2.4.3

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

$x = 2 i, - 2 i$

Step 12

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 12.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 12.2

Add $3$ to both sides of the equation.

$x = 3$

Step 13

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 13.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 13.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 14

The final solution is all the values that make $(x^{2} + 4) (x - 3) (x + 1) = 0$ true.

$x = 2 i, - 2 i, 3, - 1$

Step 15