設 P(x) = 2x^3 - 5x^2 + x + 8
(b) 當P(2x+1) 除以x - 1時的餘數為何?
maths因式定理
2011-05-28 10:53 pm
回答 (4)
2011-05-28 11:07 pm
✔ 最佳答案
P(x) = Q(x) (x - 3) + RP(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
2011-06-02 9:32 pm
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
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
2011-05-28 11:41 pm
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
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
2011-05-28 11:14 pm
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
(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
參考: miraco
收錄日期: 2021-04-26 14:07:25
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