Question

if P(x)=x^3-3x^2+3x-4,find P(2)+P(-2)+P(0)

Answer 1

just substitute 2 in P(x) we get P(2) as 8-12+6-4 =-2. similarly P(-2) is -8-12-6+4 and P(0) as 0-3(0)+3(0)-4. so P(2)+P(-2)+P(0)=-2-22-4=-28.

Answer 2

P(x) = x³ - 3x² + 3x - 4

P(2) = (2)³ - 3(2)² + 3(2) - 4
P(2) = 8 - 3*4 + 3*2 - 4
P(2) = 8 - 12 + 6 - 4
P(2) = -2

P(-2) = (-2)³ - 3(-2)² + 3(-2) - 4
P(-2) = -8 -3*4 - 6 - 4
P(-2) = -8 - 12 - 6 - 4
P(-2) = -30

P(0) = (0)³ - 3(0)² + 3(0) - 4
P(0) = 0 - 0 + 0 - 4
P(0) = -4

P(2) + P(-2) + P(0)
= (-2) + (-30) + (-4)
= -36