多項式!!!!!!!!!!!!!

2006-11-07 2:19 am
8x³ + mx² -25x+6 可被4x-1整除

m的值?

(完極列式)

回答 (2)

2006-11-07 2:51 am
✔ 最佳答案
8x³ + mx² -25x+6 可被4x-1整除
即4x-1=0
x=1/4

8(1/4)³ + m(1/4)² -25(1/4) + 6 = 0
8(1/64) + m(1/16) - 25/4 + 6 = 0
2/16 + m/16 - 100/16 + 96/16 = 0
2 + m - 100 + 96 = 0
m = 2

counter check
put m = 2 into the equation
8x³ + 2x² -25x+6
= (4x-1)(2x² + x -6)
= (4x-1)(2x - 3)(x+2)
2006-11-07 2:56 am
設F(x)=8x³+mx²-25x+6。
因為8x³+mx²-25x+6 可被4x-1整除,
F(1/4)=0
8x³+mx²-25x+6=0
8(1/4)³+(1/4)²m-25(1/4)+6=0
8/64+m/16-25/4+96=0
2+m-100+96=0
m=2
參考: Calculation


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