Precalculus Examples

$f (x) = x^{3} + 7 x^{2} + 14 x + 8$

Step 1

Set $x^{3} + 7 x^{2} + 14 x + 8$ equal to $0$ .

$x^{3} + 7 x^{2} + 14 x + 8 = 0$

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.

$x^{3} + 8 + 7 x^{2} + 14 x = 0$

Step 2.1.2

Rewrite $8$ as $2^{3}$ .

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

Step 2.1.3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 2$ .

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

Step 2.1.4

Simplify.

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Step 2.1.4.1

Multiply $2$ by $- 1$ .

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

Step 2.1.4.2

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

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

Step 2.1.5

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

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Step 2.1.5.1

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

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

Step 2.1.5.2

Factor $7 x$ out of $14 x$ .

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

Step 2.1.5.3

Factor $7 x$ out of $7 x (x) + 7 x (2)$ .

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

Step 2.1.6

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

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Step 2.1.6.1

Factor $x + 2$ out of $7 x (x + 2)$ .

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

Step 2.1.6.2

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

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

Step 2.1.7

Add $- 2 x$ and $7 x$ .

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

Step 2.1.8

Factor.

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Step 2.1.8.1

Factor $x^{2} + 5 x + 4$ using the AC method.

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Step 2.1.8.1.1

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 $4$ and whose sum is $5$ .

$1, 4$

Step 2.1.8.1.2

Write the factored form using these integers.

$(x + 2) ((x + 1) (x + 4)) = 0$

Step 2.1.8.2

Remove unnecessary parentheses.

$(x + 2) (x + 1) (x + 4) = 0$

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 $0$ .

$x + 2 = 0$

$x + 1 = 0$

$x + 4 = 0$

Step 2.3

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

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Step 2.3.1

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

$x + 2 = 0$

Step 2.3.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.4

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

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Step 2.4.1

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

$x + 1 = 0$

Step 2.4.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.5

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 2.5.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.6

The final solution is all the values that make $(x + 2) (x + 1) (x + 4) = 0$ true.

$x = - 2, - 1, - 4$

Step 3