Precalculus Examples

$x^{5} - x^{4} + 7 x^{3} - 7 x^{2} + 12 x - 12$

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.

${(1)}^{5} - {(1)}^{4} + 7 {(1)}^{3} - 7 {(1)}^{2} + 12 (1) - 12$

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

One to any power is one.

$1 - {(1)}^{4} + 7 {(1)}^{3} - 7 {(1)}^{2} + 12 (1) - 12$

Step 4.1.2

One to any power is one.

$1 - 1 \cdot 1 + 7 {(1)}^{3} - 7 {(1)}^{2} + 12 (1) - 12$

Step 4.1.3

Multiply $- 1$ by $1$ .

$1 - 1 + 7 {(1)}^{3} - 7 {(1)}^{2} + 12 (1) - 12$

Step 4.1.4

One to any power is one.

$1 - 1 + 7 \cdot 1 - 7 {(1)}^{2} + 12 (1) - 12$

Step 4.1.5

Multiply $7$ by $1$ .

$1 - 1 + 7 - 7 {(1)}^{2} + 12 (1) - 12$

Step 4.1.6

One to any power is one.

$1 - 1 + 7 - 7 \cdot 1 + 12 (1) - 12$

Step 4.1.7

Multiply $- 7$ by $1$ .

$1 - 1 + 7 - 7 + 12 (1) - 12$

Step 4.1.8

Multiply $12$ by $1$ .

$1 - 1 + 7 - 7 + 12 - 12$

$1 - 1 + 7 - 7 + 12 - 12$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $1$ from $1$ .

$0 + 7 - 7 + 12 - 12$

Step 4.2.2

Add $0$ and $7$ .

$7 - 7 + 12 - 12$

Step 4.2.3

Subtract $7$ from $7$ .

$0 + 12 - 12$

Step 4.2.4

Add $0$ and $12$ .

$12 - 12$

Step 4.2.5

Subtract $12$ from $12$ .

$0$

$0$

$0$

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.

$\frac{x^{5} - x^{4} + 7 x^{3} - 7 x^{2} + 12 x - 12}{x - 1}$

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.

$1$	$1$	$- 1$	$7$	$- 7$	$12$	$- 12$

Step 6.2

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

$1$	$1$	$- 1$	$7$	$- 7$	$12$	$- 12$

	$1$

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 $(- 1)$ .

$1$	$1$	$- 1$	$7$	$- 7$	$12$	$- 12$
		$1$
	$1$

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.

$1$	$1$	$- 1$	$7$	$- 7$	$12$	$- 12$
		$1$
	$1$	$0$

Step 6.5

Multiply the newest entry in the result $(0)$ by the divisor $(1)$ and place the result of $(0)$ under the next term in the dividend $(7)$ .

$1$	$1$	$- 1$	$7$	$- 7$	$12$	$- 12$
		$1$	$0$
	$1$	$0$

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.

$1$	$1$	$- 1$	$7$	$- 7$	$12$	$- 12$
		$1$	$0$
	$1$	$0$	$7$

Step 6.7

Multiply the newest entry in the result $(7)$ by the divisor $(1)$ and place the result of $(7)$ under the next term in the dividend $(- 7)$ .

$1$	$1$	$- 1$	$7$	$- 7$	$12$	$- 12$
		$1$	$0$	$7$
	$1$	$0$	$7$

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.

$1$	$1$	$- 1$	$7$	$- 7$	$12$	$- 12$
		$1$	$0$	$7$
	$1$	$0$	$7$	$0$

Step 6.9

Multiply the newest entry in the result $(0)$ by the divisor $(1)$ and place the result of $(0)$ under the next term in the dividend $(12)$ .

$1$	$1$	$- 1$	$7$	$- 7$	$12$	$- 12$
		$1$	$0$	$7$	$0$
	$1$	$0$	$7$	$0$

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.

$1$	$1$	$- 1$	$7$	$- 7$	$12$	$- 12$
		$1$	$0$	$7$	$0$
	$1$	$0$	$7$	$0$	$12$

Step 6.11

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 $(- 12)$ .

$1$	$1$	$- 1$	$7$	$- 7$	$12$	$- 12$
		$1$	$0$	$7$	$0$	$12$
	$1$	$0$	$7$	$0$	$12$

Step 6.12

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

$1$	$1$	$- 1$	$7$	$- 7$	$12$	$- 12$
		$1$	$0$	$7$	$0$	$12$
	$1$	$0$	$7$	$0$	$12$	$0$

Step 6.13

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

$1 x^{4} + 0 x^{3} + 7 x^{2} + 0 x + 12$

Step 6.14

Simplify the quotient polynomial.

$x^{4} + 7 x^{2} + 12$

$x^{4} + 7 x^{2} + 12$

Step 7

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

$(x - 1) {(x^{2})}^{2} + 7 x^{2} + 12$

Step 8

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

$(x - 1) u^{2} + 7 u + 12$

Step 9

Factor $u^{2} + 7 u + 12$ using the AC method.

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Step 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 $12$ and whose sum is $7$ .

$3, 4$

Step 9.2

Write the factored form using these integers.

$(x - 1) + (u + 3) (u + 4)$

$(x - 1) (u + 3) (u + 4)$

Step 10

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

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

Step 11

Factor the left side of the equation.

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Step 11.1

Regroup terms.

$x^{5} + 7 x^{3} + 12 x - x^{4} - 7 x^{2} - 12 = 0$

Step 11.2

Factor $x$ out of $x^{5} + 7 x^{3} + 12 x$ .

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Step 11.2.1

Factor $x$ out of $x^{5}$ .

$x \cdot x^{4} + 7 x^{3} + 12 x - x^{4} - 7 x^{2} - 12 = 0$

Step 11.2.2

Factor $x$ out of $7 x^{3}$ .

$x \cdot x^{4} + x (7 x^{2}) + 12 x - x^{4} - 7 x^{2} - 12 = 0$

Step 11.2.3

Factor $x$ out of $12 x$ .

$x \cdot x^{4} + x (7 x^{2}) + x \cdot 12 - x^{4} - 7 x^{2} - 12 = 0$

Step 11.2.4

Factor $x$ out of $x \cdot x^{4} + x (7 x^{2})$ .

$x (x^{4} + 7 x^{2}) + x \cdot 12 - x^{4} - 7 x^{2} - 12 = 0$

Step 11.2.5

Factor $x$ out of $x (x^{4} + 7 x^{2}) + x \cdot 12$ .

$x (x^{4} + 7 x^{2} + 12) - x^{4} - 7 x^{2} - 12 = 0$

$x (x^{4} + 7 x^{2} + 12) - x^{4} - 7 x^{2} - 12 = 0$

Step 11.3

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

$x ({(x^{2})}^{2} + 7 x^{2} + 12) - x^{4} - 7 x^{2} - 12 = 0$

Step 11.4

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

$x (u^{2} + 7 u + 12) - x^{4} - 7 x^{2} - 12 = 0$

Step 11.5

Factor $u^{2} + 7 u + 12$ using the AC method.

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Step 11.5.1

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 $7$ .

$3, 4$

Step 11.5.2

Write the factored form using these integers.

$x ((u + 3) (u + 4)) - x^{4} - 7 x^{2} - 12 = 0$

$x ((u + 3) (u + 4)) - x^{4} - 7 x^{2} - 12 = 0$

Step 11.6

Factor.

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Step 11.6.1

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

$x ((x^{2} + 3) (x^{2} + 4)) - x^{4} - 7 x^{2} - 12 = 0$

Step 11.6.2

Remove unnecessary parentheses.

$x (x^{2} + 3) (x^{2} + 4) - x^{4} - 7 x^{2} - 12 = 0$

$x (x^{2} + 3) (x^{2} + 4) - x^{4} - 7 x^{2} - 12 = 0$

Step 11.7

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

$x (x^{2} + 3) (x^{2} + 4) - {(x^{2})}^{2} - 7 x^{2} - 12 = 0$

Step 11.8

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

$x (x^{2} + 3) (x^{2} + 4) - u^{2} - 7 u - 12 = 0$

Step 11.9

Factor by grouping.

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Step 11.9.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 12 = 12$ and whose sum is $b = - 7$ .

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Step 11.9.1.1

Factor $- 7$ out of $- 7 u$ .

$x (x^{2} + 3) (x^{2} + 4) - u^{2} - 7 u - 12 = 0$

Step 11.9.1.2

Rewrite $- 7$ as $- 3$ plus $- 4$

$x (x^{2} + 3) (x^{2} + 4) - u^{2} + (- 3 - 4) u - 12 = 0$

Step 11.9.1.3

Apply the distributive property.

$x (x^{2} + 3) (x^{2} + 4) - u^{2} - 3 u - 4 u - 12 = 0$

$x (x^{2} + 3) (x^{2} + 4) - u^{2} - 3 u - 4 u - 12 = 0$

Step 11.9.2

Factor out the greatest common factor from each group.

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Step 11.9.2.1

Group the first two terms and the last two terms.

$x (x^{2} + 3) (x^{2} + 4) - u^{2} - 3 u - 4 u - 12 = 0$

Step 11.9.2.2

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

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

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

Step 11.9.3

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

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

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

Step 11.10

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

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

Step 11.11

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

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Step 11.11.1

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

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

Step 11.11.2

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

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

Step 11.11.3

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

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

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

Step 11.12

Apply the distributive property.

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

Step 11.13

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 11.13.1

Multiply $x$ by $x^{2}$ .

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Step 11.13.1.1

Raise $x$ to the power of $1$ .

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

Step 11.13.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

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

Step 11.13.2

Add $1$ and $2$ .

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

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

Step 11.14

Move $3$ to the left of $x$ .

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

Step 11.15

Reorder terms.

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

Step 11.16

Factor.

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Step 11.16.1

Rewrite $x^{3} - x^{2} + 3 x - 3$ in a factored form.

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Step 11.16.1.1

Factor out the greatest common factor from each group.

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Step 11.16.1.1.1

Group the first two terms and the last two terms.

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

Step 11.16.1.1.2

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

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

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

Step 11.16.1.2

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

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

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

Step 11.16.2

Remove unnecessary parentheses.

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

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

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

Step 12

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

$x^{2} + 4 = 0$

$x - 1 = 0$

$x^{2} + 3 = 0$

Step 13

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

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Step 13.1

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

$x^{2} + 4 = 0$

Step 13.2

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

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Step 13.2.1

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

Step 13.2.2

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

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

Step 13.2.3

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

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Step 13.2.3.1

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

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

Step 13.2.3.2

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

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

Step 13.2.3.3

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

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

Step 13.2.3.4

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

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

Step 13.2.3.5

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

$x = \pm i \cdot 2$

Step 13.2.3.6

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

$x = \pm 2 i$

$x = \pm 2 i$

Step 13.2.4

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

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Step 13.2.4.1

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

$x = 2 i$

Step 13.2.4.2

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

$x = - 2 i$

Step 13.2.4.3

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

$x = 2 i, - 2 i$

$x = 2 i, - 2 i$

$x = 2 i, - 2 i$

$x = 2 i, - 2 i$

Step 14

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

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Step 14.1

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

$x - 1 = 0$

Step 14.2

Add $1$ to both sides of the equation.

$x = 1$

$x = 1$

Step 15

Set $x^{2} + 3$ equal to $0$ and solve for $x$ .

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Step 15.1

Set $x^{2} + 3$ equal to $0$ .

$x^{2} + 3 = 0$

Step 15.2

Solve $x^{2} + 3 = 0$ for $x$ .

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Step 15.2.1

Subtract $3$ from both sides of the equation.

$x^{2} = - 3$

Step 15.2.2

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

$x = \pm \sqrt{- 3}$

Step 15.2.3

Simplify $\pm \sqrt{- 3}$ .

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Step 15.2.3.1

Rewrite $- 3$ as $- 1 (3)$ .

$x = \pm \sqrt{- 1 (3)}$

Step 15.2.3.2

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{3}$

Step 15.2.3.3

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

$x = \pm i \sqrt{3}$

$x = \pm i \sqrt{3}$

Step 15.2.4

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

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Step 15.2.4.1

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

$x = i \sqrt{3}$

Step 15.2.4.2

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

$x = - i \sqrt{3}$

Step 15.2.4.3

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

$x = i \sqrt{3}, - i \sqrt{3}$

$x = i \sqrt{3}, - i \sqrt{3}$

$x = i \sqrt{3}, - i \sqrt{3}$

$x = i \sqrt{3}, - i \sqrt{3}$

Step 16

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

$x = 2 i, - 2 i, 1, i \sqrt{3}, - i \sqrt{3}$

Step 17

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$