Given P(x) = x^3 - 3x^2 - 5x + 10
a evaluate P9x) for each integer value of x from -3 through 5
b. Prove that 5 is an upper bound on the zeros of P(x)
(b)
P(x) = x^3 - 3x^2 - 5x + 10
P'(x)=3x^2-6x-5
P"(x)=6x-6
P(5)=125-75-25+10=35>0
P'(5)=75-30-5=40>0
P"(5)=6x-6=30-6=24>0 ∀x>5
=> P'(x)>P'(5)>0 ∀x>5
=> P(x)>P(5)>0 ∀x>5
=> 5 is the upper-bound on the zeroes of P(x).