Precalculus Examples

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

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.

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

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

One to any power is one.

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

Step 4.1.2

One to any power is one.

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

Step 4.1.3

Multiply $- 1$ by $1$ .

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

Step 4.1.4

One to any power is one.

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

Step 4.1.5

Multiply $- 1$ by $1$ .

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

Step 4.1.6

One to any power is one.

$1 - 1 - 1 + 1 - 12 \cdot 1 + 12$

Step 4.1.7

Multiply $- 12$ by $1$ .

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

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

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $1$ from $1$ .

$0 - 1 + 1 - 12 + 12$

Step 4.2.2

Subtract $1$ from $0$ .

$- 1 + 1 - 12 + 12$

Step 4.2.3

Add $- 1$ and $1$ .

$0 - 12 + 12$

Step 4.2.4

Subtract $12$ from $0$ .

$- 12 + 12$

Step 4.2.5

Add $- 12$ and $12$ .

$0$

$0$

$0$

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.

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

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.

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

Step 6.2

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

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

	$1$

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 $(- 1)$ .

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

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.

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

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 $(- 1)$ .

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

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.

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

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 $(1)$ .

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

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.

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

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 $(- 12)$ .

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

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.

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

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 $(12)$ .

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

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.

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

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.

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

Step 6.14

Simplify the quotient polynomial.

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

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

Step 7

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

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

Step 8

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

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

Step 9

Factor $u^{2} - u - 12$ 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 $- 1$ .

$- 4, 3$

Step 9.2

Write the factored form using these integers.

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

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

Step 10

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

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

Step 11

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

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

Step 12

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

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

Step 13

Factor the left side of the equation.

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Step 13.1

Regroup terms.

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

Step 13.2

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

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Step 13.2.1

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

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

Step 13.2.2

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

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

Step 13.2.3

Factor $x$ out of $- 12 x$ .

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

Step 13.2.4

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

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

Step 13.2.5

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

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

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

Step 13.3

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

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

Step 13.4

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

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

Step 13.5

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

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Step 13.5.1

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 $- 1$ .

$- 4, 3$

Step 13.5.2

Write the factored form using these integers.

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

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

Step 13.6

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

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

Step 13.7

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

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

Step 13.8

Factor.

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Step 13.8.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

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

Step 13.8.2

Remove unnecessary parentheses.

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

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

Step 13.9

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

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

Step 13.10

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

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

Step 13.11

Factor by grouping.

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Step 13.11.1

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 $b = 1$ .

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Step 13.11.1.1

Multiply by $1$ .

$x (x + 2) (x - 2) (x^{2} + 3) - u^{2} + 1 u + 12 = 0$

Step 13.11.1.2

Rewrite $1$ as $- 3$ plus $4$

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

Step 13.11.1.3

Apply the distributive property.

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

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

Step 13.11.2

Factor out the greatest common factor from each group.

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Step 13.11.2.1

Group the first two terms and the last two terms.

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

Step 13.11.2.2

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

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

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

Step 13.11.3

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

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

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

Step 13.12

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

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

Step 13.13

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

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

Step 13.14

Factor.

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Step 13.14.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

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

Step 13.14.2

Remove unnecessary parentheses.

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

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

Step 13.15

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

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Step 13.15.1

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

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

Step 13.15.2

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

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

Step 13.15.3

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

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

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

Step 13.16

Apply the distributive property.

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

Step 13.17

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

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Step 13.17.1

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

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Step 13.17.1.1

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

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

Step 13.17.1.2

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

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

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

Step 13.17.2

Add $1$ and $2$ .

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

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

Step 13.18

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

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

Step 13.19

Reorder terms.

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

Step 13.20

Factor.

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Step 13.20.1

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

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Step 13.20.1.1

Factor out the greatest common factor from each group.

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Step 13.20.1.1.1

Group the first two terms and the last two terms.

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

Step 13.20.1.1.2

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

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

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

Step 13.20.1.2

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

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

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

Step 13.20.2

Remove unnecessary parentheses.

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

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

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

Step 14

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

$x + 2 = 0$

$x - 2 = 0$

$x - 1 = 0$

$x^{2} + 3 = 0$

Step 15

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 15.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 15.2

Subtract $2$ from both sides of the equation.

$x = - 2$

$x = - 2$

Step 16

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 16.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 16.2

Add $2$ to both sides of the equation.

$x = 2$

$x = 2$

Step 17

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

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Step 17.1

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

$x - 1 = 0$

Step 17.2

Add $1$ to both sides of the equation.

$x = 1$

$x = 1$

Step 18

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

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Step 18.1

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

$x^{2} + 3 = 0$

Step 18.2

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

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Step 18.2.1

Subtract $3$ from both sides of the equation.

$x^{2} = - 3$

Step 18.2.2

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

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

Step 18.2.3

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

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Step 18.2.3.1

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

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

Step 18.2.3.2

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

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

Step 18.2.3.3

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

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

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

Step 18.2.4

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

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Step 18.2.4.1

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

$x = i \sqrt{3}$

Step 18.2.4.2

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

$x = - i \sqrt{3}$

Step 18.2.4.3

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 19

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

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

Step 20

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