Find the Roots (Zeros) f(t)=3t(t-3)(t+4)

$f (t) = 3 t (t - 3) (t + 4)$

Step 1

Set $3 t (t - 3) (t + 4)$ equal to $0$ .

$3 t (t - 3) (t + 4) = 0$

Step 2

Solve for $x$ .

Divide each term in $3 t (t - 3) (t + 4) = 0$ by $3$ and simplify.

Divide each term in $3 t (t - 3) (t + 4) = 0$ by $3$ .

$\frac{3 t (t - 3) (t + 4)}{3} = \frac{0}{3}$

Simplify the left side.

Simplify terms.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 t (t - 3) (t + 4)}{3} = \frac{0}{3}$

Divide $(t (t - 3)) (t + 4)$ by $1$ .

$(t (t - 3)) (t + 4) = \frac{0}{3}$

Apply the distributive property.

$(t \cdot t + t \cdot - 3) (t + 4) = \frac{0}{3}$

Simplify the expression.

Multiply $t$ by $t$ .

$(t^{2} + t \cdot - 3) (t + 4) = \frac{0}{3}$

Move $- 3$ to the left of $t$ .

$(t^{2} - 3 \cdot t) (t + 4) = \frac{0}{3}$

Expand $(t^{2} - 3 t) (t + 4)$ using the FOIL Method.

Apply the distributive property.

$t^{2} (t + 4) - 3 t (t + 4) = \frac{0}{3}$

Apply the distributive property.

$t^{2} t + t^{2} \cdot 4 - 3 t (t + 4) = \frac{0}{3}$

Apply the distributive property.

$t^{2} t + t^{2} \cdot 4 - 3 t \cdot t - 3 t \cdot 4 = \frac{0}{3}$

Simplify and combine like terms.

Simplify each term.

Multiply $t^{2}$ by $t$ by adding the exponents.

Multiply $t^{2}$ by $t$ .

Raise $t$ to the power of $1$ .

$t^{2} t^{1} + t^{2} \cdot 4 - 3 t \cdot t - 3 t \cdot 4 = \frac{0}{3}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$t^{2 + 1} + t^{2} \cdot 4 - 3 t \cdot t - 3 t \cdot 4 = \frac{0}{3}$

Add $2$ and $1$ .

$t^{3} + t^{2} \cdot 4 - 3 t \cdot t - 3 t \cdot 4 = \frac{0}{3}$

Move $4$ to the left of $t^{2}$ .

$t^{3} + 4 \cdot t^{2} - 3 t \cdot t - 3 t \cdot 4 = \frac{0}{3}$

Multiply $t$ by $t$ by adding the exponents.

Move $t$ .

$t^{3} + 4 t^{2} - 3 (t \cdot t) - 3 t \cdot 4 = \frac{0}{3}$

Multiply $t$ by $t$ .

$t^{3} + 4 t^{2} - 3 t^{2} - 3 t \cdot 4 = \frac{0}{3}$

Multiply $4$ by $- 3$ .

$t^{3} + 4 t^{2} - 3 t^{2} - 12 t = \frac{0}{3}$

Subtract $3 t^{2}$ from $4 t^{2}$ .

$t^{3} + t^{2} - 12 t = \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

$t^{3} + t^{2} - 12 t = 0$

Factor the left side of the equation.

Factor $t$ out of $t^{3} + t^{2} - 12 t$ .

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

$t \cdot t^{2} + t^{2} - 12 t = 0$

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

$t \cdot t^{2} + t \cdot t - 12 t = 0$

Factor $t$ out of $- 12 t$ .

$t \cdot t^{2} + t \cdot t + t \cdot - 12 = 0$

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

$t (t^{2} + t) + t \cdot - 12 = 0$

Factor $t$ out of $t (t^{2} + t) + t \cdot - 12$ .

$t (t^{2} + t - 12) = 0$

Factor.

Factor $t^{2} + t - 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

$t ((t - 3) (t + 4)) = 0$

Remove unnecessary parentheses.

$t (t - 3) (t + 4) = 0$

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

$t = 0$

$t - 3 = 0$

$t + 4 = 0$

Set $t$ equal to $0$ .

$t = 0$

Set $t - 3$ equal to $0$ and solve for $t$ .

Set $t - 3$ equal to $0$ .

$t - 3 = 0$

Add $3$ to both sides of the equation.

$t = 3$

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

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

$t + 4 = 0$

Subtract $4$ from both sides of the equation.

$t = - 4$

The final solution is all the values that make $t (t - 3) (t + 4) = 0$ true.

$t = 0, 3, - 4$

Step 3

Find the Roots (Zeros) f(t)=3t(t-3)(t+4)

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