Precalculus Examples

Given that $f (x) = x^{2} + 5 x - 14$ and $g (x) = x - 2$ ; find $f (x) \cdot g (x)$ and express the result in standard form.

Step 1

Replace the function designators in $f (x) \cdot (g (x))$ with the actual functions.

$(x^{2} + 5 x - 14) \cdot (x - 2)$

Step 2

Simplify $(x^{2} + 5 x - 14) \cdot (x - 2)$ .

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Step 2.1

Expand $(x^{2} + 5 x - 14) (x - 2)$ by multiplying each term in the first expression by each term in the second expression.

$x^{2} x + x^{2} \cdot - 2 + 5 x \cdot x + 5 x \cdot - 2 - 14 x - 14 \cdot - 2$

Step 2.2

Simplify terms.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 2.2.1.1.1

Multiply $x^{2}$ by $x$ .

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Step 2.2.1.1.1.1

Raise $x$ to the power of $1$ .

$x^{2} x^{1} + x^{2} \cdot - 2 + 5 x \cdot x + 5 x \cdot - 2 - 14 x - 14 \cdot - 2$

Step 2.2.1.1.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x^{2 + 1} + x^{2} \cdot - 2 + 5 x \cdot x + 5 x \cdot - 2 - 14 x - 14 \cdot - 2$

Step 2.2.1.1.2

Add $2$ and $1$ .

$x^{3} + x^{2} \cdot - 2 + 5 x \cdot x + 5 x \cdot - 2 - 14 x - 14 \cdot - 2$

Step 2.2.1.2

Move $- 2$ to the left of $x^{2}$ .

$x^{3} - 2 \cdot x^{2} + 5 x \cdot x + 5 x \cdot - 2 - 14 x - 14 \cdot - 2$

Step 2.2.1.3

Multiply $x$ by $x$ by adding the exponents.

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Step 2.2.1.3.1

Move $x$ .

$x^{3} - 2 x^{2} + 5 (x \cdot x) + 5 x \cdot - 2 - 14 x - 14 \cdot - 2$

Step 2.2.1.3.2

Multiply $x$ by $x$ .

$x^{3} - 2 x^{2} + 5 x^{2} + 5 x \cdot - 2 - 14 x - 14 \cdot - 2$

Step 2.2.1.4

Multiply $- 2$ by $5$ .

$x^{3} - 2 x^{2} + 5 x^{2} - 10 x - 14 x - 14 \cdot - 2$

Step 2.2.1.5

Multiply $- 14$ by $- 2$ .

$x^{3} - 2 x^{2} + 5 x^{2} - 10 x - 14 x + 28$

Step 2.2.2

Simplify by adding terms.

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Step 2.2.2.1

Add $- 2 x^{2}$ and $5 x^{2}$ .

$x^{3} + 3 x^{2} - 10 x - 14 x + 28$

Step 2.2.2.2

Subtract $14 x$ from $- 10 x$ .

$x^{3} + 3 x^{2} - 24 x + 28$