F.4 MATHS

a)當f(x)除以x-3時,所得餘數是-22 , 餘式定理 :f(3) = - 22
(3 + a)(3 + b)(3 - 4) + 8 = -22
(3 + a)(3 + b) = 30b)因 a > b 都是正整數 , (3 + a) , (3 + b) 都是正整數 ,
得(3 + a) > (3 + b) > 3
(3 + a)(3 + b) = 30 的唯一可能是 = 6 x 5(3 + a) = 6 , a = 3
(3 + b) = 5 , b = 2c)x+4 是 f(x) +k = (x+3)(x+2)(x-4)+8 +k 的因式 ,
餘式定理 :
f(- 4) +k = 0(-4+3)(-4+2)(-4-4)+8 +k = 0
k = 8f(x) +k = (x+3) (x+2)(x-4)+8 +8= (x+4 - 1) (x+2)(x-4)+8 +8= (x+4)(x+2)(x-4) - (x+2)(x-4) + 16= (x+4)(x+2)(x-4) - (x+4 -2)(x-4) + 16= (x+4)(x+2)(x-4) - (x+4)(x-4) + 2(x-4) + 16= (x+4)(x+2)(x-4) - (x+4)(x-4) + 2(x+4 - 8) + 16= (x+4)(x+2)(x-4) - (x+4)(x-4) + 2(x+4)= (x+4) [(x+2)(x-4) - (x-4) + 2]= (x+4) (x² - 3x - 2)
2)a/(2x-1) - b/(x+3) + 2/(2x² + 5x - 3) ≡ 9 / [(2x-1)(x+3)][a(x+3) - b(2x-1)] / [(2x-1)(x+3)] + 2/[(2x-1)(x+3)] ≡ 9 / [(2x-1)(x+3)][a(x+3) - b(2x-1) + 2] / [(2x-1)(x+3)] ≡ 9 / [(2x-1)(x+3)][a(x+3) - b(2x-1) + 2] ≡ 9 ax + 3a - 2bx + b + 2 ≡ 9(a - 2b)x + 3a+b+2 ≡ 0x + 9比較係數 :a - 2b = 0.....(1)
3a+b+2 = 9....(2)(2)*2 + (1) :7a + 4 = 18a = 2
b = 1