Find the Roots (Zeros) h(t)=-3t^3(t+2)(t+5)

$h (t) = - 3 t^{3} (t + 2) (t + 5)$

Step 1

Set $- 3 t^{3} (t + 2) (t + 5)$ equal to $0$ .

$- 3 t^{3} (t + 2) (t + 5) = 0$

Step 2

Solve for $x$ .

Divide each term in $- 3 t^{3} (t + 2) (t + 5) = 0$ by $- 3$ and simplify.

Divide each term in $- 3 t^{3} (t + 2) (t + 5) = 0$ by $- 3$ .

$\frac{- 3 t^{3} (t + 2) (t + 5)}{- 3} = \frac{0}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 t^{3} (t + 2) (t + 5)}{- 3} = \frac{0}{- 3}$

Divide $(t^{3} (t + 2)) (t + 5)$ by $1$ .

$(t^{3} (t + 2)) (t + 5) = \frac{0}{- 3}$

Apply the distributive property.

$(t^{3} t + t^{3} \cdot 2) (t + 5) = \frac{0}{- 3}$

Multiply $t^{3}$ by $t$ by adding the exponents.

Multiply $t^{3}$ by $t$ .

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

$(t^{3} t^{1} + t^{3} \cdot 2) (t + 5) = \frac{0}{- 3}$

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

$(t^{3 + 1} + t^{3} \cdot 2) (t + 5) = \frac{0}{- 3}$

Add $3$ and $1$ .

$(t^{4} + t^{3} \cdot 2) (t + 5) = \frac{0}{- 3}$

Move $2$ to the left of $t^{3}$ .

$(t^{4} + 2 \cdot t^{3}) (t + 5) = \frac{0}{- 3}$

Expand $(t^{4} + 2 t^{3}) (t + 5)$ using the FOIL Method.

Apply the distributive property.

$t^{4} (t + 5) + 2 t^{3} (t + 5) = \frac{0}{- 3}$

Apply the distributive property.

$t^{4} t + t^{4} \cdot 5 + 2 t^{3} (t + 5) = \frac{0}{- 3}$

Apply the distributive property.

$t^{4} t + t^{4} \cdot 5 + 2 t^{3} t + 2 t^{3} \cdot 5 = \frac{0}{- 3}$

Simplify and combine like terms.

Simplify each term.

Multiply $t^{4}$ by $t$ by adding the exponents.

Multiply $t^{4}$ by $t$ .

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

$t^{4} t^{1} + t^{4} \cdot 5 + 2 t^{3} t + 2 t^{3} \cdot 5 = \frac{0}{- 3}$

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

$t^{4 + 1} + t^{4} \cdot 5 + 2 t^{3} t + 2 t^{3} \cdot 5 = \frac{0}{- 3}$

Add $4$ and $1$ .

$t^{5} + t^{4} \cdot 5 + 2 t^{3} t + 2 t^{3} \cdot 5 = \frac{0}{- 3}$

Move $5$ to the left of $t^{4}$ .

$t^{5} + 5 \cdot t^{4} + 2 t^{3} t + 2 t^{3} \cdot 5 = \frac{0}{- 3}$

Multiply $t^{3}$ by $t$ by adding the exponents.

Move $t$ .

$t^{5} + 5 t^{4} + 2 (t \cdot t^{3}) + 2 t^{3} \cdot 5 = \frac{0}{- 3}$

Multiply $t$ by $t^{3}$ .

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

$t^{5} + 5 t^{4} + 2 (t^{1} t^{3}) + 2 t^{3} \cdot 5 = \frac{0}{- 3}$

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

$t^{5} + 5 t^{4} + 2 t^{1 + 3} + 2 t^{3} \cdot 5 = \frac{0}{- 3}$

Add $1$ and $3$ .

$t^{5} + 5 t^{4} + 2 t^{4} + 2 t^{3} \cdot 5 = \frac{0}{- 3}$

Multiply $5$ by $2$ .

$t^{5} + 5 t^{4} + 2 t^{4} + 10 t^{3} = \frac{0}{- 3}$

Add $5 t^{4}$ and $2 t^{4}$ .

$t^{5} + 7 t^{4} + 10 t^{3} = \frac{0}{- 3}$

Simplify the right side.

Divide $0$ by $- 3$ .

$t^{5} + 7 t^{4} + 10 t^{3} = 0$

Factor the left side of the equation.

Factor $t^{3}$ out of $t^{5} + 7 t^{4} + 10 t^{3}$ .

Factor $t^{3}$ out of $t^{5}$ .

$t^{3} t^{2} + 7 t^{4} + 10 t^{3} = 0$

Factor $t^{3}$ out of $7 t^{4}$ .

$t^{3} t^{2} + t^{3} (7 t) + 10 t^{3} = 0$

Factor $t^{3}$ out of $10 t^{3}$ .

$t^{3} t^{2} + t^{3} (7 t) + t^{3} \cdot 10 = 0$

Factor $t^{3}$ out of $t^{3} t^{2} + t^{3} (7 t)$ .

$t^{3} (t^{2} + 7 t) + t^{3} \cdot 10 = 0$

Factor $t^{3}$ out of $t^{3} (t^{2} + 7 t) + t^{3} \cdot 10$ .

$t^{3} (t^{2} + 7 t + 10) = 0$

Factor.

Factor $t^{2} + 7 t + 10$ 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 $10$ and whose sum is $7$ .

$2, 5$

Write the factored form using these integers.

$t^{3} ((t + 2) (t + 5)) = 0$

Remove unnecessary parentheses.

$t^{3} (t + 2) (t + 5) = 0$

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

$t^{3} = 0$

$t + 2 = 0$

$t + 5 = 0$

Set $t^{3}$ equal to $0$ and solve for $t$ .

Set $t^{3}$ equal to $0$ .

$t^{3} = 0$

Solve $t^{3} = 0$ for $t$ .

Take the cube root of both sides of the equation to eliminate the exponent on the left side.

$t = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

Rewrite $0$ as $0^{3}$ .

$t = \sqrt[3]{0^{3}}$

Pull terms out from under the radical, assuming real numbers.

$t = 0$

Set $t + 2$ equal to $0$ and solve for $t$ .

Set $t + 2$ equal to $0$ .

$t + 2 = 0$

Subtract $2$ from both sides of the equation.

$t = - 2$

Set $t + 5$ equal to $0$ and solve for $t$ .

Set $t + 5$ equal to $0$ .

$t + 5 = 0$

Subtract $5$ from both sides of the equation.

$t = - 5$

The final solution is all the values that make $t^{3} (t + 2) (t + 5) = 0$ true.

$t = 0, - 2, - 5$

Step 3

Find the Roots (Zeros) h(t)=-3t^3(t+2)(t+5)

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