H(t)=t^2+4t+3 what are the zeroes of the function? Write the smaller t first, and then the larger t second

t = - 3, t = - 1

Step-by-step explanation:

Given

h(t) = t² + 4t + 3

To find the zeros equate h(t) to zero, that is

t² + 4t + 3 = 0

Consider the factors of the constant term (+ 3) which sum to give the coefficient of the t- term

The factors are + 3 and + 1, since

3 × 1 = 3 and 3 + 1 = 4, thus

t² + 4t + 3 = (t + 3)(t + 1) = 0

Equate each factor to 0 and solve for t

t + 3 = 0 ⇒ t = - 3

t + 1 = 0 ⇒ t = - 1

smaller zero is t = - 3

larger zero is t = - 1

step-by-step explanation: