maths因式定理

P(x) = Q(x) (x - 3) + R

P(2x + 1)

= Q(2x + 1)(2x - 2) + R

= 2(x - 1)Q(2x + 1) + R

So, the remainder of P(2x + 1) divided by x - 1 is

P(3) = 20

letP(x) = 2x³ - 5x² + x + 8 (b)
當P(2x+1) 除以x - 1時的餘數為何?
let f(x) = P(2x + 1) f(x) = 2(2x + 1)³ - 5(2x + 1)² + (2x + 1) + 8 P(2x+1) 除以x - 1時的餘數 = f(x) 除以x - 1時的餘數 = f(1) = 2(2*1 + 1)³ - 5(2*1 + 1)² + (2*1 + 1) + 8 = 2(3)³ - 5(3)² + (3) + 8 = 20

用餘式定理便可
P(2x+1)=2(2x+1)³-5(2x+1)²+(2x+1)+8
P(2x+1)除以(x-1)的餘數即代x=1入P(2x+1)內
∴餘數=P(3)=2(27)-5(9)+3+8=20

設 P(x) = 2x³ - 5x² + x + 8
(b) 當P(2x+1) 除以x - 1時的餘數為何?


設 f(x) = P(2x + 1)
f(x) = 2(2x + 1)³ - 5(2x + 1)² + (2x + 1) + 8

P(2x+1) 除以x - 1時的餘數
= f(x) 除以x - 1時的餘數
= f(1)
= 2(2*1 + 1)³ - 5(2*1 + 1)² + (2*1 + 1) + 8
= 2(3)³ - 5(3)² + (3) + 8
= 20