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標題:

3條因式定理(急問)

發問:

1.解下列方程 x^3-8x^2+17x-10=0 2.若x+2為P(x)=x^3+4x^2-x+k的因式, (a.)求k的值 (b.)由此,解方程P(x)=0(如需要時,答案須以根式表示) 3.設f(x)=6x^3-17x^2+x+10 (a.)證明f(x)能被2x-5整除 (b.)由此,把f(x)因式分解 急問!!?麻煩大家幫幫我

最佳解答:

1. f(x)=x^3-8x^2+17x-10 ---(1) f(1)=1-8+17-10=0 Let f(x)=(x-1)(ax^2 +bx +c)=ax^3 +(b-a)x^2 +(c-b)x -c ---(2) By comparing eqt. (1) & (2), we found that a=1,b=-7,c=10 So, f(x)=(x-1)(x^2 -7x +10)=(x-1)(x-2)(x-5) 2a. P(x)=x^3+4x^2-x+k P(-2)=0 So, -8 +16 +2 +k=0 and k= -10 2b. P(x)=x^3+4x^2-x-10 ---(1) Let P(x)=(x+2)(ax^2 +bx +c)=ax^3 +(b+2a)x^2 +(c+2b)x +2c ---(2) By comparing eqt. (1) & (2), we found that a=1,b=2,c=(-5) So, P(x)=(x+2)(x^2 +2x -5) To simply x^2 +2x -5 =0 x= [-2 +/- sqrt.(4+20)]/2 = (-2 +/- sqrt.24)/2 = -1 +/- sqrt.6 x= (-1 +sqrt.6) or (-1 -sqrt.6) Therefore, P(x)=(x+2)(x+1 - sqrt.6)(x+1 + sqrt.6) 3a. f(x)=6x^3-17x^2+x+10 f(5/2)= 6(125/8) - 17(25/4) +5/2 +10 = 375/4 - 425/4 +10/4 +40/4 =0 Therefore, f(x) is divisible by (2x-5) 3b. f(x)=6x^3-17x^2+x+10 ---(1) Let f(x)=(2x-5)(ax^2 +bx +c)=2ax^3 +(2b-5a)x^2 +(2c-5b)x -5c ---(2) By comparing eqt. (1) & (2), we found that a=3,b=(-1),c=(-2) So, f(x)=(2x-5)(3x^2 -x -2)=(2x-5)(3x+2)(x-1)

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