Polynomial functions question please help..?

Method 1 :

P(t) = -t³ + 12t² - 21t + 10
P'(t) = -3t² + 24t - 21
P"(t) = -6t + 24

When P'(t) = 0 :
-3t² + 24t - 21 = 0
t² - 8t + 7 = 0
(t - 1)(t - 7) = 0
t = 1 or t = 7

When t = 1: P'(t) = 0 and P"(t) = -6×1 + 24 = 18 > 0
Hence, minimum P(t) at t = 1

When t = 7: P'(t) = 0 and P"(t) = -6×7 + 24 = -18 < 0
Hence maximum P(t) at t = 7
Maximum P(t) = -7³ + 12×7² - 21×7 + 10 = 108

Ans: The company make their maximum profit of $108 000 in the 7th year.


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Method 2 :

P(t) = -t³ + 12t² - 21t + 10

When the maximum profit = $108 000, P(t) = 108

Maximum P(t) :
-t³ + 12t² - 21t + 10 = 108
t³ - 12t² + 21t + 98 = 0

Let f(t) = t³ - 12t² + 21t + 98
f(7) = 7³ - 12×7² + 21×7 + 98 = 0
Hence, (t - 7) is a factor of f(x).

t³ - 12t² + 21t + 98 = 0
t²(t - 7) + 7t² - 12t² + 21t + 98 = 0
t²(t - 7) - 5t² + 21t + 98 = 0
t²(t - 7) - 5t(t - 7) - 35t + 21t + 98 = 0
t²(t - 7) - 5t(t - 7) - 14t + 98 = 0
t²(t - 7) - 5t(t - 7) - 14(t - 7) = 0
(t - 7)(t² - 5t - 14) = 0
(t - 7)(t - 7)(t + 2) = 0
t = 7 or t = 7 or t = -2 (rejected)

Ans: The company make their maximum profit of $108 000 in the 7th year.

P(t) = –t³ + 12t² – 21t + 10
to get max, differentiate and set equal to zero
P' = –3t² + 24t – 21 = 0
t² – 8t + 7 = 0
(t – 7)(t – 1) = 0
t = 1, 7 years

to find which is the answer
try t = 1
–t³ + 12t² – 21t + 10 = 108
–1 + 12 – 21 + 10 = 108
0 = 108
not this one

try t = 7
–t³ + 12t² – 21t + 10 = 108
–343 + 588 – 147 + 10 = 108
108 = 108
this is the answer, year 7

P(t) = -t^3 + 12t^2 - 21t + 10

dP/dt = - 3t^2 + 24t - 21

For max/min profit, dP/dt = 0

- 3t^2 + 24t - 21 = 0

t^2 - 8t + 7 = 0

(t - 7)(t - 1) = 0

t = 1, 7

d^2P/dt^2 = - 6t + 24

For t = 1, d^2P/dt^2 = 18 > 0 ====> minimum

For t = 7, d^2P/dt^2 = - 18 < 0 ====> maximum

After 7 years