What is (n^2)/(n+4)-(n+2)/(n^2-16) ?

The solution to (n^2)/(n+4)-(n+2)/(n^2-16) is (n^3-4n^2-n-2)/((n+4)(n-4))

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simplify (n^2)/(n+4)-(n+2)/(n^2-16)

simplify

\frac{n ^{2}}{n + 4} - \frac{n + 2}{n ^{2} - 16}

\frac{n ^{3} - 4 n ^{2} - n - 2}{( n + 4 ) ( n - 4 )}

Solution steps

\frac{n ^{2}}{n + 4} - \frac{n + 2}{n ^{2} - 16}

Least Common Multiplier of

n + 4, n^{2} - 16 : (n + 4) (n - 4)

Adjust Fractions based on the LCM

= \frac{n ^{2} ( n - 4 )}{( n + 4 ) ( n - 4 )} - \frac{n + 2}{( n + 4 ) ( n - 4 )}

Apply the fraction rule:

\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}

= \frac{n ^{2} ( n - 4 ) - ( n + 2 )}{( n + 4 ) ( n - 4 )}

Simplify

n^{2} (n - 4) - (n + 2) : n^{3} - 4 n^{2} - n - 2

= \frac{n ^{3} - 4 n ^{2} - n - 2}{( n + 4 ) ( n - 4 )}

4 n^{9} + 12 n

23 - \frac{x}{5} > \frac{x}{2} - 12

3 + 2

\frac{5}{- 5 + 4 i}

- 6 (2 x + 1) - 3 (x - 4) = - 15 x + 1

What is (n^2)/(n+4)-(n+2)/(n^2-16) ?
The solution to (n^2)/(n+4)-(n+2)/(n^2-16) is (n^3-4n^2-n-2)/((n+4)(n-4))