Solve by Factoring 12n^2+48n=-n^3-64

$12 n^{2} + 48 n = - n^{3} - 64$

Move all the expressions to the left side of the equation.

Add $n^{3}$ to both sides of the equation.

$12 n^{2} + 48 n + n^{3} = - 64$

Add $64$ to both sides of the equation.

$12 n^{2} + 48 n + n^{3} + 64 = 0$

Reorder terms.

$n^{3} + 12 n^{2} + 48 n + 64 = 0$

Rewrite $n^{3} + 12 n^{2} + 48 n + 64$ in a factored form.

Regroup terms.

$n^{3} + 64 + 12 n^{2} + 48 n = 0$

Rewrite $64$ as $4^{3}$ .

$n^{3} + 4^{3} + 12 n^{2} + 48 n = 0$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = n$ and $b = 4$ .

$(n + 4) (n^{2} - n \cdot 4 + 4^{2}) + 12 n^{2} + 48 n = 0$

Simplify.

Multiply $4$ by $- 1$ .

$(n + 4) (n^{2} - 4 n + 4^{2}) + 12 n^{2} + 48 n = 0$

Raise $4$ to the power of $2$ .

$(n + 4) (n^{2} - 4 n + 16) + 12 n^{2} + 48 n = 0$

Factor $12 n$ out of $12 n^{2} + 48 n$ .

Factor $12 n$ out of $12 n^{2}$ .

$(n + 4) (n^{2} - 4 n + 16) + 12 n (n) + 48 n = 0$

Factor $12 n$ out of $48 n$ .

$(n + 4) (n^{2} - 4 n + 16) + 12 n (n) + 12 n (4) = 0$

Factor $12 n$ out of $12 n (n) + 12 n (4)$ .

$(n + 4) (n^{2} - 4 n + 16) + 12 n (n + 4) = 0$

Factor $n + 4$ out of $(n + 4) (n^{2} - 4 n + 16) + 12 n (n + 4)$ .

Factor $n + 4$ out of $12 n (n + 4)$ .

$(n + 4) (n^{2} - 4 n + 16) + (n + 4) (12 n) = 0$

Factor $n + 4$ out of $(n + 4) (n^{2} - 4 n + 16) + (n + 4) (12 n)$ .

$(n + 4) (n^{2} - 4 n + 16 + 12 n) = 0$

Add $- 4 n$ and $12 n$ .

$(n + 4) (n^{2} + 8 n + 16) = 0$

Factor using the perfect square rule.

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

$(n + 4) (n^{2} + 8 n + 4^{2}) = 0$

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

$8 n = 2 \cdot n \cdot 4$

Rewrite the polynomial.

$(n + 4) (n^{2} + 2 \cdot n \cdot 4 + 4^{2}) = 0$

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

$(n + 4) {(n + 4)}^{2} = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$n + 4 = 0$

${(n + 4)}^{2} = 0$

Set $n + 4$ equal to $0$ and solve for $n$ .

Set $n + 4$ equal to $0$ .

$n + 4 = 0$

Subtract $4$ from both sides of the equation.

$n = - 4$

The final solution is all the values that make $(n + 4) {(n + 4)}^{2} = 0$ true.

$n = - 4$

Solve by Factoring 12n^2+48n=-n^3-64

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