What is (3n)/(n^2+4n+4)+7/(n^2+6n+8) ?

The solution to (3n)/(n^2+4n+4)+7/(n^2+6n+8) is (3n^2+19n+14)/((n+2)^2(n+4))

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simplify (3n)/(n^2+4n+4)+7/(n^2+6n+8)

simplify

\frac{3 n}{n ^{2} + 4 n + 4} + \frac{7}{n ^{2} + 6 n + 8}

\frac{3 n ^{2} + 19 n + 14}{( n + 2 ) ^{2} ( n + 4 )}

Solution steps

\frac{3 n}{n ^{2} + 4 n + 4} + \frac{7}{n ^{2} + 6 n + 8}

Least Common Multiplier of

n^{2} + 4 n + 4, n^{2} + 6 n + 8 : (n + 2)^{2} (n + 4)

Adjust Fractions based on the LCM

= \frac{3 n ( n + 4 )}{( n + 2 ) ^{2} ( n + 4 )} + \frac{7 ( n + 2 )}{( n + 2 ) ^{2} ( n + 4 )}

Apply the fraction rule:

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

= \frac{3 n ( n + 4 ) + 7 ( n + 2 )}{( n + 2 ) ^{2} ( n + 4 )}

Simplify

3 n (n + 4) + 7 (n + 2) : 3 n^{2} + 19 n + 14

= \frac{3 n ^{2} + 19 n + 14}{( n + 2 ) ^{2} ( n + 4 )}

- \frac{1}{3} (- 3)

49 n^{2} + 168 n + 144

\frac{2 x + 1}{3} = - 3

4 x > - 12

P = - 2^{3} + 7 (- 2)^{2} - 2 (- 2) + 40

What is (3n)/(n^2+4n+4)+7/(n^2+6n+8) ?
The solution to (3n)/(n^2+4n+4)+7/(n^2+6n+8) is (3n^2+19n+14)/((n+2)^2(n+4))