Zeros of Polynomial Functions...2

Section 2.5
Question 14



h(t) = (t - 1)^2 - (t + 1)^2

h(t) = (t - 1)t - 1) - [(t + 1)(t + 1)]

h(t) = t^2 - 2t + 1 - [t^2 + 2t + 1]

h(t) = t^2 - 2t + 1 - t^2 - 2t - 1

h(t) = -2t - 2t

h(t) = -4t

Let h(t) = 0.

0 = -4t

0/-4 = t

0 = t

The only zero is 0 found at the origin.

 

perfect

 

As long as the concept is understood the rest falls into place.