Precalculus Examples

If $f (x) = x^{3} + x^{2} - 22 x - 40$ and $f (- 2) = 0$ ; then find all of the zeros of $f (x)$ algebraically.

Step 1

Evaluate $f (- 2)$ .

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Step 1.1

Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = {(- 2)}^{3} + {(- 2)}^{2} - 22 \cdot - 2 - 40$

Step 1.2

Simplify ${(- 2)}^{3} + {(- 2)}^{2} - 22 \cdot - 2 - 40$ .

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Step 1.2.1

Remove parentheses.

${(- 2)}^{3} + {(- 2)}^{2} - 22 \cdot - 2 - 40$

Step 1.2.2

Simplify each term.

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Step 1.2.2.1

Raise $- 2$ to the power of $3$ .

$- 8 + {(- 2)}^{2} - 22 \cdot - 2 - 40$

Step 1.2.2.2

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

$- 8 + 4 - 22 \cdot - 2 - 40$

Step 1.2.2.3

Multiply $- 22$ by $- 2$ .

$- 8 + 4 + 44 - 40$

Step 1.2.3

Simplify by adding and subtracting.

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Step 1.2.3.1

Add $- 8$ and $4$ .

$- 4 + 44 - 40$

Step 1.2.3.2

Add $- 4$ and $44$ .

$40 - 40$

Step 1.2.3.3

Subtract $40$ from $40$ .

$0$

Step 2

Solve $0 = 0$ .

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Step 2.1

Since $0 = 0$ , the equation will always be true.

Always true

Step 3

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$