Question

Solve the equation. 5t^3 +40t^2 = -80t
a. t= 0 and t= 4
b. t= -4 and t=0
c. t= -4 and t= 5
d. t= 4

Answer 1

Answer:

Option 'A' is true.

Step-by-step explanation:

Given the equation

$5t^3\:+40t^2\:=\:-80t$

Add 80 to both sides

$5t^3+40t^2+80t=-80t+80t$

Simplify

$5t^3+40t^2+80t=0$

as 5t³ + 40t² + 80t = 5t (t + 4)²

5t (t + 4)² = 0

Using the zero factor principle

if ab=0, then a=0 or b=0 (or both a=0 and b=0)

$t=0$        or        $t+4=0$

solving

$t + 4 = 0$

subtract 4 from both sides

$t+4-4=0-4$

Simplify

$t=-4$

Therefore, the solution to the equation is:

$t=0,\:t=-4$

Hence, option 'A' is true.