設f(x)=8X^3-10X^2+X+1(a)(i)求f(1)(ii)因式分解f(x)(b)在△ABC中,BC=a ㎝,AB=c ㎝,a及c是二次方程X^2-8X+6=0的根,而cos(A+C)是方程f(x)=0的最小的根,求AC?

(a) (i) f(1) = 8(1)³-10(1)²+1+1 = 0

(ii) f(x) 的一根 = x-1 - - - - 從(i)
設 f(x) = (x-1)(px²+qx+r)
=> 8x³-10x²+x+1 = (x-1)(px²+qx+r)
=> 8x³-10x²+x+1 = px³+(q-p)x²+(r-q)x-r
=> p=8, r=-1, q-p= -10, r-q = 1
=> p=8, r=-1, q-8= -10
=> p=8, r=-1, q= -2
∴ 8x³-10x²+x+1 = (x-1)(8x²-2x-1) = (x-1)(4x+1)(2x-1)

(b) 2x²-8x+6=0
=> 2(x²-4x+3)=0
=> (x-1)(x-3)=0
=> x=1,3
∴ a,c = 1,3

f(x)=8x³-10x²+x+1 = (x-1)(4x+1)(2x-1)
∴ f(x) =0 的最小根 = -1/4
=> cos(A+C) = -1/4
=> cos(180°-B) = -1/4 . . . . . . [ ∵ ABC 是 △ ]
=> - cos B = -1/4 . . . . . . [ ∵ cos(180°-B) = - cos B ]
=> cos B =1/4

b²= a²+c² -2ac cos B
　= 1²+3² - (1)(3)(1/4)
　= 10 - (3/4)
　= 37/4
∴ AC = b = (√37) / 2