How do you divide $\frac{2 v + 5}{4 v^{4} - v^{2}} \div \frac{2 v + 5}{4 v^{2} - 1}$ $\frac { 2v + 5} { 4v ^ { 4} - v ^ { 2} } \div \frac { 2v + 5} { 4v ^ { 2} - 1}$ ?

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How do you divide $\frac{2 v + 5}{4 v^{4} - v^{2}} \div \frac{2 v + 5}{4 v^{2} - 1}$ $\frac { 2v + 5} { 4v ^ { 4} - v ^ { 2} } \div \frac { 2v + 5} { 4v ^ { 2} - 1}$ ?

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Answer 1 · 2017-07-05T04:10:48

$\frac{4 v^{2} - 1}{4 v^{4} - v^{2}}$ $(4v^2-1)/(4v^4-v^2)$

Explanation:

First, we write the division in fraction form (notice: you don't have to do this if you're familiar with division by two fractions):
$\frac{\frac{2 v + 5}{4 v^{4} - v^{2}}}{\frac{2 v + 5}{4 v^{2} - 1}}$ $((2v+5)/(4v^4-v^2))/((2v+5)/(4v^2-1))$

Simplify:
$\frac{\frac{2 v + 5}{4 v^{4} - v^{2}}}{\frac{2 v + 5}{4 v^{2} - 1}}$ $((2v+5)/(4v^4-v^2))/((2v+5)/(4v^2-1))$ = $\frac{(4 v^{2} - 1) (2 v + 5)}{(4 v^{4} - v^{2}) (2 v + 5)}$ $((4v^2-1)(2v+5))/((4v^4-v^2)(2v+5))$

Cancel out like terms:
$\frac{\frac{2 v + 5}{4 v^{4} - v^{2}}}{\frac{2 v + 5}{4 v^{2} - 1}}$ $((2v+5)/(4v^4-v^2))/((2v+5)/(4v^2-1))$ = $((4v^2-1)cancel((2v+5)))/((4v^4-v^2)cancel((2v+5)))$

And you get $(4v^2-1)/(4v^4-v^2)$

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How do you divide $\frac{2 v + 5}{4 v^{4} - v^{2}} \div \frac{2 v + 5}{4 v^{2} - 1}$ $\frac { 2v + 5} { 4v ^ { 4} - v ^ { 2} } \div \frac { 2v + 5} { 4v ^ { 2} - 1}$ ?

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Explanation:

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Explanation:

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How do you divide $\frac{2 v + 5}{4 v^{4} - v^{2}} \div \frac{2 v + 5}{4 v^{2} - 1}$ $\frac { 2v + 5} { 4v ^ { 4} - v ^ { 2} } \div \frac { 2v + 5} { 4v ^ { 2} - 1}$ ?