Precalculus Examples

$x^{3} - x^{2} - 30 x = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor $x$ out of $x^{3} - x^{2} - 30 x$ .

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Step 1.1.1

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

$x \cdot x^{2} - x^{2} - 30 x = 0$

Step 1.1.2

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

$x \cdot x^{2} + x (- x) - 30 x = 0$

Step 1.1.3

Factor $x$ out of $- 30 x$ .

$x \cdot x^{2} + x (- x) + x \cdot - 30 = 0$

Step 1.1.4

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

$x (x^{2} - x) + x \cdot - 30 = 0$

Step 1.1.5

Factor $x$ out of $x (x^{2} - x) + x \cdot - 30$ .

$x (x^{2} - x - 30) = 0$

Step 1.2

Factor.

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Step 1.2.1

Factor $x^{2} - x - 30$ using the AC method.

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Step 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 $- 30$ and whose sum is $- 1$ .

$- 6, 5$

Step 1.2.1.2

Write the factored form using these integers.

$x ((x - 6) (x + 5)) = 0$

Step 1.2.2

Remove unnecessary parentheses.

$x (x - 6) (x + 5) = 0$

Step 2

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

$x = 0$

$x - 6 = 0$

$x + 5 = 0$

Step 3

Set $x$ equal to $0$ .

$x = 0$

Step 4

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 4.2

Add $6$ to both sides of the equation.

$x = 6$

Step 5

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 5.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 6

The final solution is all the values that make $x (x - 6) (x + 5) = 0$ true.

$x = 0, 6, - 5$