Simplify (a^2+16)/(a-4)+(8a)/(4-a)

$\frac{a^{2} + 16}{a - 4} + \frac{8 a}{4 - a}$

Simplify terms.

Rewrite $4$ as $- 1 (- 4)$ .

$\frac{a^{2} + 16}{a - 4} + \frac{8 a}{- 1 (- 4) - a}$

Factor $- 1$ out of $- a$ .

$\frac{a^{2} + 16}{a - 4} + \frac{8 a}{- 1 (- 4) - (a)}$

Factor $- 1$ out of $- 1 (- 4) - (a)$ .

$\frac{a^{2} + 16}{a - 4} + \frac{8 a}{- 1 (- 4 + a)}$

Simplify the expression.

Move a negative from the denominator of $\frac{8 a}{- 1 (- 4 + a)}$ to the numerator.

$\frac{a^{2} + 16}{a - 4} + \frac{- (8 a)}{- 4 + a}$

Reorder terms.

$\frac{a^{2} + 16}{a - 4} + \frac{- (8 a)}{a - 4}$

Combine the numerators over the common denominator.

$\frac{a^{2} + 16 - (8 a)}{a - 4}$

Factor using the perfect square rule.

Rearrange terms.

$\frac{a^{2} - 1 \cdot 8 a + 16}{a - 4}$

Rewrite $16$ as $4^{2}$ .

$\frac{a^{2} - 1 \cdot 8 a + 4^{2}}{a - 4}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$1 \cdot 8 a = 2 \cdot a \cdot 4$

Rewrite the polynomial.

$\frac{a^{2} - 2 \cdot a \cdot 4 + 4^{2}}{a - 4}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = a$ and $b = 4$ .

$\frac{{(a - 4)}^{2}}{a - 4}$

Cancel the common factor of ${(a - 4)}^{2}$ and $a - 4$ .

Factor $a - 4$ out of ${(a - 4)}^{2}$ .

$\frac{(a - 4) (a - 4)}{a - 4}$

Cancel the common factors.

Multiply by $1$ .

$\frac{(a - 4) (a - 4)}{(a - 4) \cdot 1}$

Cancel the common factor.

$\frac{(a - 4) (a - 4)}{(a - 4) \cdot 1}$

Rewrite the expression.

$\frac{a - 4}{1}$

Divide $a - 4$ by $1$ .

$a - 4$