Precalculus Examples

Given that $f (x) = x^{2} - 12 x + 27$ and $g (x) = x - 9$ ; 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.

$(x^{2} - 12 x + 27) \cdot (x - 9)$

Step 2

Simplify $(x^{2} - 12 x + 27) \cdot (x - 9)$ .

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Step 2.1

Expand $(x^{2} - 12 x + 27) (x - 9)$ by multiplying each term in the first expression by each term in the second expression.

$x^{2} x + x^{2} \cdot - 9 - 12 x \cdot x - 12 x \cdot - 9 + 27 x + 27 \cdot - 9$

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 $1$ .

$x^{2} x^{1} + x^{2} \cdot - 9 - 12 x \cdot x - 12 x \cdot - 9 + 27 x + 27 \cdot - 9$

Step 2.2.1.1.1.2

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

$x^{2 + 1} + x^{2} \cdot - 9 - 12 x \cdot x - 12 x \cdot - 9 + 27 x + 27 \cdot - 9$

Step 2.2.1.1.2

Add $2$ and $1$ .

$x^{3} + x^{2} \cdot - 9 - 12 x \cdot x - 12 x \cdot - 9 + 27 x + 27 \cdot - 9$

Step 2.2.1.2

Move $- 9$ to the left of $x^{2}$ .

$x^{3} - 9 \cdot x^{2} - 12 x \cdot x - 12 x \cdot - 9 + 27 x + 27 \cdot - 9$

Step 2.2.1.3

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

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Step 2.2.1.3.1

Move $x$ .

$x^{3} - 9 x^{2} - 12 (x \cdot x) - 12 x \cdot - 9 + 27 x + 27 \cdot - 9$

Step 2.2.1.3.2

Multiply $x$ by $x$ .

$x^{3} - 9 x^{2} - 12 x^{2} - 12 x \cdot - 9 + 27 x + 27 \cdot - 9$

Step 2.2.1.4

Multiply $- 9$ by $- 12$ .

$x^{3} - 9 x^{2} - 12 x^{2} + 108 x + 27 x + 27 \cdot - 9$

Step 2.2.1.5

Multiply $27$ by $- 9$ .

$x^{3} - 9 x^{2} - 12 x^{2} + 108 x + 27 x - 243$

Step 2.2.2

Simplify by adding terms.

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Step 2.2.2.1

Subtract $12 x^{2}$ from $- 9 x^{2}$ .

$x^{3} - 21 x^{2} + 108 x + 27 x - 243$

Step 2.2.2.2

Add $108 x$ and $27 x$ .

$x^{3} - 21 x^{2} + 135 x - 243$