Solve for t t^3+18t^2+81t=0

$t^{3} + 18 t^{2} + 81 t = 0$

Factor the left side of the equation.

Factor $t$ out of $t^{3} + 18 t^{2} + 81 t$ .

Factor $t$ out of $t^{3}$ .

$t \cdot t^{2} + 18 t^{2} + 81 t = 0$

Factor $t$ out of $18 t^{2}$ .

$t \cdot t^{2} + t (18 t) + 81 t = 0$

Factor $t$ out of $81 t$ .

$t \cdot t^{2} + t (18 t) + t \cdot 81 = 0$

Factor $t$ out of $t \cdot t^{2} + t (18 t)$ .

$t (t^{2} + 18 t) + t \cdot 81 = 0$

Factor $t$ out of $t (t^{2} + 18 t) + t \cdot 81$ .

$t (t^{2} + 18 t + 81) = 0$

Factor using the perfect square rule.

Rewrite $81$ as $9^{2}$ .

$t (t^{2} + 18 t + 9^{2}) = 0$

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

$18 t = 2 \cdot t \cdot 9$

Rewrite the polynomial.

$t (t^{2} + 2 \cdot t \cdot 9 + 9^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = t$ and $b = 9$ .

$t {(t + 9)}^{2} = 0$

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

$t = 0$

${(t + 9)}^{2} = 0$

Set $t$ equal to $0$ .

$t = 0$

Set ${(t + 9)}^{2}$ equal to $0$ and solve for $t$ .

Set ${(t + 9)}^{2}$ equal to $0$ .

${(t + 9)}^{2} = 0$

Solve ${(t + 9)}^{2} = 0$ for $t$ .

Set the $t + 9$ equal to $0$ .

$t + 9 = 0$

Subtract $9$ from both sides of the equation.

$t = - 9$

The final solution is all the values that make $t {(t + 9)}^{2} = 0$ true.

$t = 0, - 9$