Find the Quotient of the Two Functions p(x)=x^3-8 , d(x)=x+2

$p (x) = x^{3} - 8$ , $d (x) = x + 2$

Step 1

Replace the function designators with the actual functions in $\frac{p (x)}{d (x)}$ .

$\frac{x^{3} - 8}{x + 2}$

Step 2

Simplify the numerator.

Rewrite $8$ as $2^{3}$ .

$\frac{x^{3} - 2^{3}}{x + 2}$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 2$ .

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

Simplify.

Move $2$ to the left of $x$ .

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

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

$\frac{(x - 2) (x^{2} + 2 x + 4)}{x + 2}$

Find the Quotient of the Two Functions p(x)=x^3-8 , d(x)=x+2

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