Simplify (3n^2+14n)/(9n^2-64)+16/(9n^2-64)

$\frac{3 n^{2} + 14 n}{9 n^{2} - 64} + \frac{16}{9 n^{2} - 64}$

Step 1

Combine the numerators over the common denominator.

$\frac{3 n^{2} + 14 n + 16}{9 n^{2} - 64}$

Step 2

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot 16 = 48$ and whose sum is $b = 14$ .

Factor $14$ out of $14 n$ .

$\frac{3 n^{2} + 14 (n) + 16}{9 n^{2} - 64}$

Rewrite $14$ as $6$ plus $8$

$\frac{3 n^{2} + (6 + 8) n + 16}{9 n^{2} - 64}$

Apply the distributive property.

$\frac{3 n^{2} + 6 n + 8 n + 16}{9 n^{2} - 64}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(3 n^{2} + 6 n) + 8 n + 16}{9 n^{2} - 64}$

Factor out the greatest common factor (GCF) from each group.

$\frac{3 n (n + 2) + 8 (n + 2)}{9 n^{2} - 64}$

Factor the polynomial by factoring out the greatest common factor, $n + 2$ .

$\frac{(n + 2) (3 n + 8)}{9 n^{2} - 64}$

Step 3

Simplify the denominator.

Rewrite $9 n^{2}$ as ${(3 n)}^{2}$ .

$\frac{(n + 2) (3 n + 8)}{{(3 n)}^{2} - 64}$

Rewrite $64$ as $8^{2}$ .

$\frac{(n + 2) (3 n + 8)}{{(3 n)}^{2} - 8^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 n$ and $b = 8$ .

$\frac{(n + 2) (3 n + 8)}{(3 n + 8) (3 n - 8)}$

Step 4

Cancel the common factor of $3 n + 8$ .

Cancel the common factor.

$\frac{(n + 2) (3 n + 8)}{(3 n + 8) (3 n - 8)}$

Rewrite the expression.

$\frac{n + 2}{3 n - 8}$

Simplify (3n^2+14n)/(9n^2-64)+16/(9n^2-64)

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