f(x) is polynomial and is divisible by (x-1). What must be a factor of f(2x+1)?
Since f(x) is divisible by (x-1), f(1) = 0
Hence, f(2(0)+1) = f(1) = 0
Therefore, f(2x+1) is divisible by x.
But I do not understand why f(1)=0 is also applicable for f(2x+1)
Please explain & show more detail steps - Many Thanks
Maths F4 - Remainder Theorem
2007-09-10 3:30 am
回答 (2)
2007-09-10 6:21 am
✔ 最佳答案
f(x) is divisible by (x-1). So f(x)=(x-1)g(x)Maybe it's clearer to use a different varible y for the equation,
f(y) = (y-1)g(y)
let y=2x+1
f(2x+1)=(2x+1-1)g(2x+1)
f(2x+1)=2x.g(2x+1)
so f(2x+1) is divisible by 2x.
2007-09-10 3:38 am
佢話f(x)divisible(x-1),
f(1)=0,f(1)divisible(1-1)=0
f(2(0)+1)=0,f(2(0)+1)divisible2(0)+1-1=0
f(1)=0,f(1)divisible(1-1)=0
f(2(0)+1)=0,f(2(0)+1)divisible2(0)+1-1=0
收錄日期: 2021-04-13 14:56:54
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