Precalculus Examples

$x^{5} - 4 x^{4} - 3 x^{3} + 22 x^{2} - 4 x - 24$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$

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.

${(- 1)}^{5} - 4 {(- 1)}^{4} - 3 {(- 1)}^{3} + 22 {(- 1)}^{2} - 4 \cdot - 1 - 24$

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 $5$ .

$- 1 - 4 {(- 1)}^{4} - 3 {(- 1)}^{3} + 22 {(- 1)}^{2} - 4 \cdot - 1 - 24$

Step 4.1.2

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

$- 1 - 4 \cdot 1 - 3 {(- 1)}^{3} + 22 {(- 1)}^{2} - 4 \cdot - 1 - 24$

Step 4.1.3

Multiply $- 4$ by $1$ .

$- 1 - 4 - 3 {(- 1)}^{3} + 22 {(- 1)}^{2} - 4 \cdot - 1 - 24$

Step 4.1.4

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

$- 1 - 4 - 3 \cdot - 1 + 22 {(- 1)}^{2} - 4 \cdot - 1 - 24$

Step 4.1.5

Multiply $- 3$ by $- 1$ .

$- 1 - 4 + 3 + 22 {(- 1)}^{2} - 4 \cdot - 1 - 24$

Step 4.1.6

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

$- 1 - 4 + 3 + 22 \cdot 1 - 4 \cdot - 1 - 24$

Step 4.1.7

Multiply $22$ by $1$ .

$- 1 - 4 + 3 + 22 - 4 \cdot - 1 - 24$

Step 4.1.8

Multiply $- 4$ by $- 1$ .

$- 1 - 4 + 3 + 22 + 4 - 24$

$- 1 - 4 + 3 + 22 + 4 - 24$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $4$ from $- 1$ .

$- 5 + 3 + 22 + 4 - 24$

Step 4.2.2

Add $- 5$ and $3$ .

$- 2 + 22 + 4 - 24$

Step 4.2.3

Add $- 2$ and $22$ .

$20 + 4 - 24$

Step 4.2.4

Add $20$ and $4$ .

$24 - 24$

Step 4.2.5

Subtract $24$ from $24$ .

$0$

$0$

$0$

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.

$\frac{x^{5} - 4 x^{4} - 3 x^{3} + 22 x^{2} - 4 x - 24}{x + 1}$

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.

$- 1$	$1$	$- 4$	$- 3$	$22$	$- 4$	$- 24$

Step 6.2

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

$- 1$	$1$	$- 4$	$- 3$	$22$	$- 4$	$- 24$

	$1$

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 $(- 4)$ .

$- 1$	$1$	$- 4$	$- 3$	$22$	$- 4$	$- 24$
		$- 1$
	$1$

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.

$- 1$	$1$	$- 4$	$- 3$	$22$	$- 4$	$- 24$
		$- 1$
	$1$	$- 5$

Step 6.5

Multiply the newest entry in the result $(- 5)$ by the divisor $(- 1)$ and place the result of $(5)$ under the next term in the dividend $(- 3)$ .

$- 1$	$1$	$- 4$	$- 3$	$22$	$- 4$	$- 24$
		$- 1$	$5$
	$1$	$- 5$

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.

$- 1$	$1$	$- 4$	$- 3$	$22$	$- 4$	$- 24$
		$- 1$	$5$
	$1$	$- 5$	$2$

Step 6.7

Multiply the newest entry in the result $(2)$ by the divisor $(- 1)$ and place the result of $(- 2)$ under the next term in the dividend $(22)$ .

$- 1$	$1$	$- 4$	$- 3$	$22$	$- 4$	$- 24$
		$- 1$	$5$	$- 2$
	$1$	$- 5$	$2$

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.

$- 1$	$1$	$- 4$	$- 3$	$22$	$- 4$	$- 24$
		$- 1$	$5$	$- 2$
	$1$	$- 5$	$2$	$20$

Step 6.9

Multiply the newest entry in the result $(20)$ by the divisor $(- 1)$ and place the result of $(- 20)$ under the next term in the dividend $(- 4)$ .

$- 1$	$1$	$- 4$	$- 3$	$22$	$- 4$	$- 24$
		$- 1$	$5$	$- 2$	$- 20$
	$1$	$- 5$	$2$	$20$

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.

$- 1$	$1$	$- 4$	$- 3$	$22$	$- 4$	$- 24$
		$- 1$	$5$	$- 2$	$- 20$
	$1$	$- 5$	$2$	$20$	$- 24$

Step 6.11

Multiply the newest entry in the result $(- 24)$ by the divisor $(- 1)$ and place the result of $(24)$ under the next term in the dividend $(- 24)$ .

$- 1$	$1$	$- 4$	$- 3$	$22$	$- 4$	$- 24$
		$- 1$	$5$	$- 2$	$- 20$	$24$
	$1$	$- 5$	$2$	$20$	$- 24$

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.

$- 1$	$1$	$- 4$	$- 3$	$22$	$- 4$	$- 24$
		$- 1$	$5$	$- 2$	$- 20$	$24$
	$1$	$- 5$	$2$	$20$	$- 24$	$0$

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.

$1 x^{4} + - 5 x^{3} + 2 x^{2} + 20 x - 24$

Step 6.14

Simplify the quotient polynomial.

$x^{4} - 5 x^{3} + 2 x^{2} + 20 x - 24$

$x^{4} - 5 x^{3} + 2 x^{2} + 20 x - 24$

Step 7

Solve the equation to find any remaining roots.

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Step 7.1

Factor the left side of the equation.

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Step 7.1.1

Regroup terms.

$- 5 x^{3} + 20 x + x^{4} + 2 x^{2} - 24 = 0$

Step 7.1.2

Factor $- 5 x$ out of $- 5 x^{3} + 20 x$ .

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Step 7.1.2.1

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

$- 5 x \cdot x^{2} + 20 x + x^{4} + 2 x^{2} - 24 = 0$

Step 7.1.2.2

Factor $- 5 x$ out of $20 x$ .

$- 5 x \cdot x^{2} - 5 x \cdot - 4 + x^{4} + 2 x^{2} - 24 = 0$

Step 7.1.2.3

Factor $- 5 x$ out of $- 5 x (x^{2}) - 5 x (- 4)$ .

$- 5 x (x^{2} - 4) + x^{4} + 2 x^{2} - 24 = 0$

$- 5 x (x^{2} - 4) + x^{4} + 2 x^{2} - 24 = 0$

Step 7.1.3

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

$- 5 x (x^{2} - 2^{2}) + x^{4} + 2 x^{2} - 24 = 0$

Step 7.1.4

Factor.

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Step 7.1.4.1

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 $b = 2$ .

$- 5 x ((x + 2) (x - 2)) + x^{4} + 2 x^{2} - 24 = 0$

Step 7.1.4.2

Remove unnecessary parentheses.

$- 5 x (x + 2) (x - 2) + x^{4} + 2 x^{2} - 24 = 0$

$- 5 x (x + 2) (x - 2) + x^{4} + 2 x^{2} - 24 = 0$

Step 7.1.5

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

$- 5 x (x + 2) (x - 2) + {(x^{2})}^{2} + 2 x^{2} - 24 = 0$

Step 7.1.6

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

$- 5 x (x + 2) (x - 2) + u^{2} + 2 u - 24 = 0$

Step 7.1.7

Factor $u^{2} + 2 u - 24$ using the AC method.

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Step 7.1.7.1

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 $- 24$ and whose sum is $2$ .

$- 4, 6$

Step 7.1.7.2

Write the factored form using these integers.

$- 5 x (x + 2) (x - 2) + (u - 4) (u + 6) = 0$

$- 5 x (x + 2) (x - 2) + (u - 4) (u + 6) = 0$

Step 7.1.8

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

$- 5 x (x + 2) (x - 2) + (x^{2} - 4) (x^{2} + 6) = 0$

Step 7.1.9

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

$- 5 x (x + 2) (x - 2) + (x^{2} - 2^{2}) (x^{2} + 6) = 0$

Step 7.1.10

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 $b = 2$ .

$- 5 x (x + 2) (x - 2) + (x + 2) (x - 2) (x^{2} + 6) = 0$

Step 7.1.11

Factor $(x + 2) (x - 2)$ out of $- 5 x (x + 2) (x - 2) + (x + 2) (x - 2) (x^{2} + 6)$ .

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Step 7.1.11.1

Factor $(x + 2) (x - 2)$ out of $- 5 x (x + 2) (x - 2)$ .

$(x + 2) (x - 2) (- 5 x) + (x + 2) (x - 2) (x^{2} + 6) = 0$

Step 7.1.11.2

Factor $(x + 2) (x - 2)$ out of $(x + 2) (x - 2) (- 5 x) + (x + 2) (x - 2) (x^{2} + 6)$ .

$(x + 2) (x - 2) (- 5 x + x^{2} + 6) = 0$

$(x + 2) (x - 2) (- 5 x + x^{2} + 6) = 0$

Step 7.1.12

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

$(x + 2) (x - 2) (- 5 u + u^{2} + 6) = 0$

Step 7.1.13

Factor $- 5 u + u^{2} + 6$ using the AC method.

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Step 7.1.13.1

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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Step 7.1.13.2

Write the factored form using these integers.

$(x + 2) (x - 2) ((u - 3) (u - 2)) = 0$

$(x + 2) (x - 2) ((u - 3) (u - 2)) = 0$

Step 7.1.14

Factor.

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Step 7.1.14.1

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

$(x + 2) (x - 2) ((x - 3) (x - 2)) = 0$

Step 7.1.14.2

Remove unnecessary parentheses.

$(x + 2) (x - 2) (x - 3) (x - 2) = 0$

$(x + 2) (x - 2) (x - 3) (x - 2) = 0$

Step 7.1.15

Combine exponents.

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Step 7.1.15.1

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

$(x + 2) ((x - 2) (x - 2)) (x - 3) = 0$

Step 7.1.15.2

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

$(x + 2) ((x - 2) (x - 2)) (x - 3) = 0$

Step 7.1.15.3

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

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

Step 7.1.15.4

Add $1$ and $1$ .

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

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

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

Step 7.2

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

$x + 2 = 0$

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

$x - 3 = 0$

Step 7.3

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

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Step 7.3.1

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

$x + 2 = 0$

Step 7.3.2

Subtract $2$ from both sides of the equation.

$x = - 2$

$x = - 2$

Step 7.4

Set ${(x - 2)}^{2}$ equal to $0$ and solve for $x$ .

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Step 7.4.1

Set ${(x - 2)}^{2}$ equal to $0$ .

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

Step 7.4.2

Solve ${(x - 2)}^{2} = 0$ for $x$ .

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Step 7.4.2.1

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

$x - 2 = 0$

Step 7.4.2.2

Add $2$ to both sides of the equation.

$x = 2$

$x = 2$

$x = 2$

Step 7.5

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

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Step 7.5.1

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

$x - 3 = 0$

Step 7.5.2

Add $3$ to both sides of the equation.

$x = 3$

$x = 3$

Step 7.6

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

$x = - 2, 2, 3$

$x = - 2, 2, 3$

Step 8

The polynomial can be written as a set of linear factors.

$(x + 1) (x + 2) (x - 2) (x - 3)$

Step 9

These are the roots (zeros) of the polynomial $x^{5} - 4 x^{4} - 3 x^{3} + 22 x^{2} - 4 x - 24$ .

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

Step 10

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