Precalculus Examples

$x^{4} + 12 x^{3} + 37 x^{2} + 12 x + 36$

Step 1

Set $x^{4} + 12 x^{3} + 37 x^{2} + 12 x + 36$ equal to $0$ .

$x^{4} + 12 x^{3} + 37 x^{2} + 12 x + 36 = 0$

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.

$12 x^{3} + 12 x + x^{4} + 37 x^{2} + 36 = 0$

Step 2.1.2

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

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Step 2.1.2.1

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

$12 x (x^{2}) + 12 x + x^{4} + 37 x^{2} + 36 = 0$

Step 2.1.2.2

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

$12 x (x^{2}) + 12 x (1) + x^{4} + 37 x^{2} + 36 = 0$

Step 2.1.2.3

Factor $12 x$ out of $12 x (x^{2}) + 12 x (1)$ .

$12 x (x^{2} + 1) + x^{4} + 37 x^{2} + 36 = 0$

Step 2.1.3

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

$12 x (x^{2} + 1) + {(x^{2})}^{2} + 37 x^{2} + 36 = 0$

Step 2.1.4

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

$12 x (x^{2} + 1) + u^{2} + 37 u + 36 = 0$

Step 2.1.5

Factor $u^{2} + 37 u + 36$ using the AC method.

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Step 2.1.5.1

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 $36$ and whose sum is $37$ .

$1, 36$

Step 2.1.5.2

Write the factored form using these integers.

$12 x (x^{2} + 1) + (u + 1) (u + 36) = 0$

Step 2.1.6

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

$12 x (x^{2} + 1) + (x^{2} + 1) (x^{2} + 36) = 0$

Step 2.1.7

Factor $x^{2} + 1$ out of $12 x (x^{2} + 1) + (x^{2} + 1) (x^{2} + 36)$ .

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Step 2.1.7.1

Factor $x^{2} + 1$ out of $12 x (x^{2} + 1)$ .

$(x^{2} + 1) (12 x) + (x^{2} + 1) (x^{2} + 36) = 0$

Step 2.1.7.2

Factor $x^{2} + 1$ out of $(x^{2} + 1) (12 x) + (x^{2} + 1) (x^{2} + 36)$ .

$(x^{2} + 1) (12 x + x^{2} + 36) = 0$

Step 2.1.8

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

$(x^{2} + 1) (12 u + u^{2} + 36) = 0$

Step 2.1.9

Factor using the perfect square rule.

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Step 2.1.9.1

Rearrange terms.

$(x^{2} + 1) (u^{2} + 12 u + 36) = 0$

Step 2.1.9.2

Rewrite $36$ as $6^{2}$ .

$(x^{2} + 1) (u^{2} + 12 u + 6^{2}) = 0$

Step 2.1.9.3

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

$12 u = 2 \cdot u \cdot 6$

Step 2.1.9.4

Rewrite the polynomial.

$(x^{2} + 1) (u^{2} + 2 \cdot u \cdot 6 + 6^{2}) = 0$

Step 2.1.9.5

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 6$ .

$(x^{2} + 1) {(u + 6)}^{2} = 0$

Step 2.1.10

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

$(x^{2} + 1) {(x + 6)}^{2} = 0$

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 $0$ .

$x^{2} + 1 = 0$

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

Step 2.3

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

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Step 2.3.1

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

$x^{2} + 1 = 0$

Step 2.3.2

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

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Step 2.3.2.1

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 2.3.2.2

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

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

Step 2.3.2.3

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

$x = \pm i$

Step 2.3.2.4

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

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Step 2.3.2.4.1

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

$x = i$

Step 2.3.2.4.2

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

$x = - i$

Step 2.3.2.4.3

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

$x = i, - i$

Step 2.4

Set ${(x + 6)}^{2}$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set ${(x + 6)}^{2}$ equal to $0$ .

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

Step 2.4.2

Solve ${(x + 6)}^{2} = 0$ for $x$ .

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Step 2.4.2.1

Set the $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 2.4.2.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 2.5

The final solution is all the values that make $(x^{2} + 1) {(x + 6)}^{2} = 0$ true.

$x = i, - i, - 6$

Step 3