Precalculus Examples

$f (x) = x^{3} + 5 x^{2} + 6 x$

Step 1

Set $x^{3} + 5 x^{2} + 6 x$ equal to $0$ .

$x^{3} + 5 x^{2} + 6 x = 0$

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

Factor $x$ out of $x^{3} + 5 x^{2} + 6 x$ .

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Step 2.1.1.1

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

$x \cdot x^{2} + 5 x^{2} + 6 x = 0$

Step 2.1.1.2

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

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

Step 2.1.1.3

Factor $x$ out of $6 x$ .

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

Step 2.1.1.4

Factor $x$ out of $x \cdot x^{2} + x (5 x)$ .

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

Step 2.1.1.5

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

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

Step 2.1.2

Factor.

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Step 2.1.2.1

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

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Step 2.1.2.1.1

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 $6$ and whose sum is $5$ .

$2, 3$

Step 2.1.2.1.2

Write the factored form using these integers.

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

Step 2.1.2.2

Remove unnecessary parentheses.

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

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 $0$ .

$x = 0$

$x + 2 = 0$

$x + 3 = 0$

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

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

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Step 2.4.1

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

$x + 2 = 0$

Step 2.4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.5

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

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Step 2.5.1

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

$x + 3 = 0$

Step 2.5.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.6

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

$x = 0, - 2, - 3$

Step 3