Precalculus Examples

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

Step 1

Set $x^{5} + 7 x^{4} + 2 x^{3} + 14 x^{2} + x + 7$ equal to $0$ .

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

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Regroup terms.

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

Step 2.1.2

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

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Step 2.1.2.1

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

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

Step 2.1.2.2

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 2.1.2.3

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 2.1.2.4

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 2.1.2.5

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 2.1.2.6

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

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

Step 2.1.3

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

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

Step 2.1.4

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 2.1.5

Factor using the perfect square rule.

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Step 2.1.5.1

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

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

Step 2.1.5.2

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 2.1.5.3

Rewrite the polynomial.

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

Step 2.1.5.4

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

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

Step 2.1.6

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

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

Step 2.1.7

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

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Step 2.1.7.1

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

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

Step 2.1.7.2

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

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

Step 2.1.7.3

Factor $7$ out of $7$ .

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

Step 2.1.7.4

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 2.1.7.5

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

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

Step 2.1.8

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

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

Step 2.1.9

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 2.1.10

Factor using the perfect square rule.

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Step 2.1.10.1

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

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

Step 2.1.10.2

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 2.1.10.3

Rewrite the polynomial.

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

Step 2.1.10.4

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

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

Step 2.1.11

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

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

Step 2.1.12

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

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Step 2.1.12.1

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

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

Step 2.1.12.2

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 2.1.12.3

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

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

Step 2.2

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

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

$x + 7 = 0$

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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Step 2.3.2.1

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

$x^{2} + 1 = 0$

Step 2.3.2.2

Solve for $x$ .

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Step 2.3.2.2.1

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 2.3.2.2.2

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

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

Step 2.3.2.2.3

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

$x = \pm i$

Step 2.3.2.2.4

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

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Step 2.3.2.2.4.1

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

$x = i$

Step 2.3.2.2.4.2

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

$x = - i$

Step 2.3.2.2.4.3

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

$x = i, - i$

Step 2.4

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

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Step 2.4.1

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

$x + 7 = 0$

Step 2.4.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 2.5

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

$x = i, - i, - 7$

Step 3