Simplify ((4p-8)/(p^3-2p^2)-(q+2)/(q^3+2q^2))*p/(2q-p)

$(\frac{4 p - 8}{p^{3} - 2 p^{2}} - \frac{q + 2}{q^{3} + 2 q^{2}}) \cdot \frac{p}{2 q - p}$

Simplify each term.

Factor $4$ out of $4 p - 8$ .

Factor $4$ out of $4 p$ .

$(\frac{4 (p) - 8}{p^{3} - 2 p^{2}} - \frac{q + 2}{q^{3} + 2 q^{2}}) \cdot \frac{p}{2 q - p}$

Factor $4$ out of $- 8$ .

$(\frac{4 p + 4 \cdot - 2}{p^{3} - 2 p^{2}} - \frac{q + 2}{q^{3} + 2 q^{2}}) \cdot \frac{p}{2 q - p}$

Factor $4$ out of $4 p + 4 \cdot - 2$ .

$(\frac{4 (p - 2)}{p^{3} - 2 p^{2}} - \frac{q + 2}{q^{3} + 2 q^{2}}) \cdot \frac{p}{2 q - p}$

Factor $p^{2}$ out of $p^{3} - 2 p^{2}$ .

Factor $p^{2}$ out of $p^{3}$ .

$(\frac{4 (p - 2)}{p^{2} p - 2 p^{2}} - \frac{q + 2}{q^{3} + 2 q^{2}}) \cdot \frac{p}{2 q - p}$

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

$(\frac{4 (p - 2)}{p^{2} p + p^{2} \cdot - 2} - \frac{q + 2}{q^{3} + 2 q^{2}}) \cdot \frac{p}{2 q - p}$

Factor $p^{2}$ out of $p^{2} p + p^{2} \cdot - 2$ .

$(\frac{4 (p - 2)}{p^{2} (p - 2)} - \frac{q + 2}{q^{3} + 2 q^{2}}) \cdot \frac{p}{2 q - p}$

Cancel the common factor of $p - 2$ .

Cancel the common factor.

$(\frac{4 (p - 2)}{p^{2} (p - 2)} - \frac{q + 2}{q^{3} + 2 q^{2}}) \cdot \frac{p}{2 q - p}$

Rewrite the expression.

$(\frac{4}{p^{2}} - \frac{q + 2}{q^{3} + 2 q^{2}}) \cdot \frac{p}{2 q - p}$

Factor $q^{2}$ out of $q^{3} + 2 q^{2}$ .

Factor $q^{2}$ out of $q^{3}$ .

$(\frac{4}{p^{2}} - \frac{q + 2}{q^{2} q + 2 q^{2}}) \cdot \frac{p}{2 q - p}$

Factor $q^{2}$ out of $2 q^{2}$ .

$(\frac{4}{p^{2}} - \frac{q + 2}{q^{2} q + q^{2} \cdot 2}) \cdot \frac{p}{2 q - p}$

Factor $q^{2}$ out of $q^{2} q + q^{2} \cdot 2$ .

$(\frac{4}{p^{2}} - \frac{q + 2}{q^{2} (q + 2)}) \cdot \frac{p}{2 q - p}$

Cancel the common factor of $q + 2$ .

Cancel the common factor.

$(\frac{4}{p^{2}} - \frac{q + 2}{q^{2} (q + 2)}) \cdot \frac{p}{2 q - p}$

Rewrite the expression.

$(\frac{4}{p^{2}} - \frac{1}{q^{2}}) \cdot \frac{p}{2 q - p}$

To write $\frac{4}{p^{2}}$ as a fraction with a common denominator, multiply by $\frac{q^{2}}{q^{2}}$ .

$(\frac{4}{p^{2}} \cdot \frac{q^{2}}{q^{2}} - \frac{1}{q^{2}}) \cdot \frac{p}{2 q - p}$

To write $- \frac{1}{q^{2}}$ as a fraction with a common denominator, multiply by $\frac{p^{2}}{p^{2}}$ .

$(\frac{4}{p^{2}} \cdot \frac{q^{2}}{q^{2}} - \frac{1}{q^{2}} \cdot \frac{p^{2}}{p^{2}}) \cdot \frac{p}{2 q - p}$

Write each expression with a common denominator of $p^{2} q^{2}$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{4}{p^{2}}$ by $\frac{q^{2}}{q^{2}}$ .

$(\frac{4 q^{2}}{p^{2} q^{2}} - \frac{1}{q^{2}} \cdot \frac{p^{2}}{p^{2}}) \cdot \frac{p}{2 q - p}$

Multiply $\frac{1}{q^{2}}$ by $\frac{p^{2}}{p^{2}}$ .

$(\frac{4 q^{2}}{p^{2} q^{2}} - \frac{p^{2}}{q^{2} p^{2}}) \cdot \frac{p}{2 q - p}$

Reorder the factors of $q^{2} p^{2}$ .

$(\frac{4 q^{2}}{p^{2} q^{2}} - \frac{p^{2}}{p^{2} q^{2}}) \cdot \frac{p}{2 q - p}$

Combine the numerators over the common denominator.

$\frac{4 q^{2} - p^{2}}{p^{2} q^{2}} \cdot \frac{p}{2 q - p}$

Simplify the numerator.

Rewrite $4 q^{2}$ as ${(2 q)}^{2}$ .

$\frac{{(2 q)}^{2} - p^{2}}{p^{2} q^{2}} \cdot \frac{p}{2 q - p}$

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

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

Reduce the expression by cancelling the common factors.

Cancel the common factor of $2 q - p$ .

Factor $2 q - p$ out of $(2 q + p) (2 q - p)$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{2 q + p}{p^{2} q^{2}} \cdot p$

Cancel the common factor of $p$ .

Factor $p$ out of $p^{2} q^{2}$ .

$\frac{2 q + p}{p (p q^{2})} \cdot p$

Cancel the common factor.

$\frac{2 q + p}{p (p q^{2})} \cdot p$

Rewrite the expression.

$\frac{2 q + p}{p q^{2}}$

Simplify ((4p-8)/(p^3-2p^2)-(q+2)/(q^3+2q^2))*p/(2q-p)

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