Solve by Factoring 2t^5+2t^4-144t^3=0

$2 t^{5} + 2 t^{4} - 144 t^{3} = 0$

Step 1

Factor $2 t^{3}$ out of $2 t^{5} + 2 t^{4} - 144 t^{3}$ .

Factor $2 t^{3}$ out of $2 t^{5}$ .

$2 t^{3} (t^{2}) + 2 t^{4} - 144 t^{3} = 0$

Factor $2 t^{3}$ out of $2 t^{4}$ .

$2 t^{3} (t^{2}) + 2 t^{3} (t) - 144 t^{3} = 0$

Factor $2 t^{3}$ out of $- 144 t^{3}$ .

$2 t^{3} (t^{2}) + 2 t^{3} (t) + 2 t^{3} (- 72) = 0$

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

$2 t^{3} (t^{2} + t) + 2 t^{3} (- 72) = 0$

Factor $2 t^{3}$ out of $2 t^{3} (t^{2} + t) + 2 t^{3} (- 72)$ .

$2 t^{3} (t^{2} + t - 72) = 0$

Step 2

Factor.

Factor $t^{2} + t - 72$ 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 $- 72$ and whose sum is $1$ .

$- 8, 9$

Write the factored form using these integers.

$2 t^{3} ((t - 8) (t + 9)) = 0$

Remove unnecessary parentheses.

$2 t^{3} (t - 8) (t + 9) = 0$

Step 3

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

$t^{3} = 0$

$t - 8 = 0$

$t + 9 = 0$

Step 4

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 specified 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.

$t = 0$

Step 5

Set $t - 8$ equal to $0$ and solve for $t$ .

Set $t - 8$ equal to $0$ .

$t - 8 = 0$

Add $8$ to both sides of the equation.

$t = 8$

Step 6

Set $t + 9$ equal to $0$ and solve for $t$ .

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

$t + 9 = 0$

Subtract $9$ from both sides of the equation.

$t = - 9$

Step 7

The final solution is all the values that make $2 t^{3} (t - 8) (t + 9) = 0$ true.

$t = 0, 8, - 9$

Solve by Factoring 2t^5+2t^4-144t^3=0

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