Precalculus Examples

$f (x) = x^{5} + 7 x^{4} + 2 x^{3} + 14 x^{2} + x + 7$

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.

$p = \pm 1, \pm 7$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 7$

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.

${(- 7)}^{5} + 7 {(- 7)}^{4} + 2 {(- 7)}^{3} + 14 {(- 7)}^{2} - 7 + 7$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 7$ is a root of the polynomial.

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Step 4.1

Remove parentheses.

${(- 7)}^{5} + 7 {(- 7)}^{4} + 2 {(- 7)}^{3} + 14 {(- 7)}^{2} - 7 + 7$

Step 4.2

Simplify each term.

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Step 4.2.1

Raise $- 7$ to the power of $5$ .

$- 16807 + 7 {(- 7)}^{4} + 2 {(- 7)}^{3} + 14 {(- 7)}^{2} - 7 + 7$

Step 4.2.2

Raise $- 7$ to the power of $4$ .

$- 16807 + 7 \cdot 2401 + 2 {(- 7)}^{3} + 14 {(- 7)}^{2} - 7 + 7$

Step 4.2.3

Multiply $7$ by $2401$ .

$- 16807 + 16807 + 2 {(- 7)}^{3} + 14 {(- 7)}^{2} - 7 + 7$

Step 4.2.4

Raise $- 7$ to the power of $3$ .

$- 16807 + 16807 + 2 \cdot - 343 + 14 {(- 7)}^{2} - 7 + 7$

Step 4.2.5

Multiply $2$ by $- 343$ .

$- 16807 + 16807 - 686 + 14 {(- 7)}^{2} - 7 + 7$

Step 4.2.6

Raise $- 7$ to the power of $2$ .

$- 16807 + 16807 - 686 + 14 \cdot 49 - 7 + 7$

Step 4.2.7

Multiply $14$ by $49$ .

$- 16807 + 16807 - 686 + 686 - 7 + 7$

$- 16807 + 16807 - 686 + 686 - 7 + 7$

Step 4.3

Simplify by adding and subtracting.

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Step 4.3.1

Add $- 16807$ and $16807$ .

$0 - 686 + 686 - 7 + 7$

Step 4.3.2

Subtract $686$ from $0$ .

$- 686 + 686 - 7 + 7$

Step 4.3.3

Add $- 686$ and $686$ .

$0 - 7 + 7$

Step 4.3.4

Subtract $7$ from $0$ .

$- 7 + 7$

Step 4.3.5

Add $- 7$ and $7$ .

$0$

$0$

$0$

Step 5

Since $- 7$ is a known root, divide the polynomial by $x + 7$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{5} + 7 x^{4} + 2 x^{3} + 14 x^{2} + x + 7}{x + 7}$

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.

$- 7$	$1$	$7$	$2$	$14$	$1$	$7$

Step 6.2

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

$- 7$	$1$	$7$	$2$	$14$	$1$	$7$

	$1$

Step 6.3

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

$- 7$	$1$	$7$	$2$	$14$	$1$	$7$
		$- 7$
	$1$

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.

$- 7$	$1$	$7$	$2$	$14$	$1$	$7$
		$- 7$
	$1$	$0$

Step 6.5

Multiply the newest entry in the result $(0)$ by the divisor $(- 7)$ and place the result of $(0)$ under the next term in the dividend $(2)$ .

$- 7$	$1$	$7$	$2$	$14$	$1$	$7$
		$- 7$	$0$
	$1$	$0$

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.

$- 7$	$1$	$7$	$2$	$14$	$1$	$7$
		$- 7$	$0$
	$1$	$0$	$2$

Step 6.7

Multiply the newest entry in the result $(2)$ by the divisor $(- 7)$ and place the result of $(- 14)$ under the next term in the dividend $(14)$ .

$- 7$	$1$	$7$	$2$	$14$	$1$	$7$
		$- 7$	$0$	$- 14$
	$1$	$0$	$2$

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.

$- 7$	$1$	$7$	$2$	$14$	$1$	$7$
		$- 7$	$0$	$- 14$
	$1$	$0$	$2$	$0$

Step 6.9

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

$- 7$	$1$	$7$	$2$	$14$	$1$	$7$
		$- 7$	$0$	$- 14$	$0$
	$1$	$0$	$2$	$0$

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.

$- 7$	$1$	$7$	$2$	$14$	$1$	$7$
		$- 7$	$0$	$- 14$	$0$
	$1$	$0$	$2$	$0$	$1$

Step 6.11

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

$- 7$	$1$	$7$	$2$	$14$	$1$	$7$
		$- 7$	$0$	$- 14$	$0$	$- 7$
	$1$	$0$	$2$	$0$	$1$

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.

$- 7$	$1$	$7$	$2$	$14$	$1$	$7$
		$- 7$	$0$	$- 14$	$0$	$- 7$
	$1$	$0$	$2$	$0$	$1$	$0$

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.

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

Step 6.14

Simplify the quotient polynomial.

$x^{4} + 2 x^{2} + 1$

$x^{4} + 2 x^{2} + 1$

Step 7

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

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

Step 8

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

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

Step 9

Factor using the perfect square rule.

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Step 9.1

Rewrite $1$ as $1^{2}$ .

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

Step 9.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 u = 2 \cdot u \cdot 1$

Step 9.3

Rewrite the polynomial.

$(x + 7) + u^{2} + 2 \cdot u \cdot 1 + 1^{2}$

Step 9.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 1$ .

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

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

Step 10

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

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

Step 11

Factor the left side of the equation.

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Step 11.1

Regroup terms.

$x^{5} + 2 x^{3} + x + 7 x^{4} + 14 x^{2} + 7 = 0$

Step 11.2

Factor $x$ out of $x^{5} + 2 x^{3} + x$ .

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Step 11.2.1

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

$x \cdot x^{4} + 2 x^{3} + x + 7 x^{4} + 14 x^{2} + 7 = 0$

Step 11.2.2

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

$x \cdot x^{4} + x (2 x^{2}) + x + 7 x^{4} + 14 x^{2} + 7 = 0$

Step 11.2.3

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

$x \cdot x^{4} + x (2 x^{2}) + x + 7 x^{4} + 14 x^{2} + 7 = 0$

Step 11.2.4

Factor $x$ out of $x^{1}$ .

$x \cdot x^{4} + x (2 x^{2}) + x \cdot 1 + 7 x^{4} + 14 x^{2} + 7 = 0$

Step 11.2.5

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

$x (x^{4} + 2 x^{2}) + x \cdot 1 + 7 x^{4} + 14 x^{2} + 7 = 0$

Step 11.2.6

Factor $x$ out of $x (x^{4} + 2 x^{2}) + x \cdot 1$ .

$x (x^{4} + 2 x^{2} + 1) + 7 x^{4} + 14 x^{2} + 7 = 0$

$x (x^{4} + 2 x^{2} + 1) + 7 x^{4} + 14 x^{2} + 7 = 0$

Step 11.3

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

$x ({(x^{2})}^{2} + 2 x^{2} + 1) + 7 x^{4} + 14 x^{2} + 7 = 0$

Step 11.4

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

$x (u^{2} + 2 u + 1) + 7 x^{4} + 14 x^{2} + 7 = 0$

Step 11.5

Factor using the perfect square rule.

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Step 11.5.1

Rewrite $1$ as $1^{2}$ .

$x (u^{2} + 2 u + 1^{2}) + 7 x^{4} + 14 x^{2} + 7 = 0$

Step 11.5.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 u = 2 \cdot u \cdot 1$

Step 11.5.3

Rewrite the polynomial.

$x (u^{2} + 2 \cdot u \cdot 1 + 1^{2}) + 7 x^{4} + 14 x^{2} + 7 = 0$

Step 11.5.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 1$ .

$x {(u + 1)}^{2} + 7 x^{4} + 14 x^{2} + 7 = 0$

$x {(u + 1)}^{2} + 7 x^{4} + 14 x^{2} + 7 = 0$

Step 11.6

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

$x {(x^{2} + 1)}^{2} + 7 x^{4} + 14 x^{2} + 7 = 0$

Step 11.7

Factor $7$ out of $7 x^{4} + 14 x^{2} + 7$ .

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Step 11.7.1

Factor $7$ out of $7 x^{4}$ .

$x {(x^{2} + 1)}^{2} + 7 (x^{4}) + 14 x^{2} + 7 = 0$

Step 11.7.2

Factor $7$ out of $14 x^{2}$ .

$x {(x^{2} + 1)}^{2} + 7 (x^{4}) + 7 (2 x^{2}) + 7 = 0$

Step 11.7.3

Factor $7$ out of $7$ .

$x {(x^{2} + 1)}^{2} + 7 (x^{4}) + 7 (2 x^{2}) + 7 (1) = 0$

Step 11.7.4

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

$x {(x^{2} + 1)}^{2} + 7 (x^{4} + 2 x^{2}) + 7 (1) = 0$

Step 11.7.5

Factor $7$ out of $7 (x^{4} + 2 x^{2}) + 7 (1)$ .

$x {(x^{2} + 1)}^{2} + 7 (x^{4} + 2 x^{2} + 1) = 0$

$x {(x^{2} + 1)}^{2} + 7 (x^{4} + 2 x^{2} + 1) = 0$

Step 11.8

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

$x {(x^{2} + 1)}^{2} + 7 ({(x^{2})}^{2} + 2 x^{2} + 1) = 0$

Step 11.9

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

$x {(x^{2} + 1)}^{2} + 7 (u^{2} + 2 u + 1) = 0$

Step 11.10

Factor using the perfect square rule.

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Step 11.10.1

Rewrite $1$ as $1^{2}$ .

$x {(x^{2} + 1)}^{2} + 7 (u^{2} + 2 u + 1^{2}) = 0$

Step 11.10.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 u = 2 \cdot u \cdot 1$

Step 11.10.3

Rewrite the polynomial.

$x {(x^{2} + 1)}^{2} + 7 (u^{2} + 2 \cdot u \cdot 1 + 1^{2}) = 0$

Step 11.10.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 1$ .

$x {(x^{2} + 1)}^{2} + 7 {(u + 1)}^{2} = 0$

$x {(x^{2} + 1)}^{2} + 7 {(u + 1)}^{2} = 0$

Step 11.11

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

$x {(x^{2} + 1)}^{2} + 7 {(x^{2} + 1)}^{2} = 0$

Step 11.12

Factor ${(x^{2} + 1)}^{2}$ out of $x {(x^{2} + 1)}^{2} + 7 {(x^{2} + 1)}^{2}$ .

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Step 11.12.1

Factor ${(x^{2} + 1)}^{2}$ out of $x {(x^{2} + 1)}^{2}$ .

${(x^{2} + 1)}^{2} x + 7 {(x^{2} + 1)}^{2} = 0$

Step 11.12.2

Factor ${(x^{2} + 1)}^{2}$ out of $7 {(x^{2} + 1)}^{2}$ .

${(x^{2} + 1)}^{2} x + {(x^{2} + 1)}^{2} \cdot 7 = 0$

Step 11.12.3

Factor ${(x^{2} + 1)}^{2}$ out of ${(x^{2} + 1)}^{2} x + {(x^{2} + 1)}^{2} \cdot 7$ .

${(x^{2} + 1)}^{2} (x + 7) = 0$

${(x^{2} + 1)}^{2} (x + 7) = 0$

${(x^{2} + 1)}^{2} (x + 7) = 0$

Step 12

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

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

$x + 7 = 0$

Step 13

Set ${(x^{2} + 1)}^{2}$ equal to $0$ and solve for $x$ .

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Step 13.1

Set ${(x^{2} + 1)}^{2}$ equal to $0$ .

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

Step 13.2

Solve ${(x^{2} + 1)}^{2} = 0$ for $x$ .

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Step 13.2.1

Set the $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Step 13.2.2

Solve for $x$ .

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Step 13.2.2.1

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 13.2.2.2

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

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

Step 13.2.2.3

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

$x = \pm i$

Step 13.2.2.4

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

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Step 13.2.2.4.1

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

$x = i$

Step 13.2.2.4.2

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

$x = - i$

Step 13.2.2.4.3

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

$x = i, - i$

$x = i, - i$

$x = i, - i$

$x = i, - i$

$x = i, - i$

Step 14

Set $x + 7$ equal to $0$ and solve for $x$ .

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Step 14.1

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Step 14.2

Subtract $7$ from both sides of the equation.

$x = - 7$

$x = - 7$

Step 15

The final solution is all the values that make ${(x^{2} + 1)}^{2} (x + 7) = 0$ true.

$x = i, - i, - 7$

Step 16

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