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標題:

MATHS~~~~FAST~~~

發問:

Polynomials:::: (i) Expand -(3x-1)(x+4) (ii) Using the result obtained in(i), expand -(3x-1)(2-x)(x-+4) THanks you

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(i) - ( 3x - 1 )( x +4) = - [ 3x(x+4) -1( x+4)] = - [3x^2 +12x -x -4] = - ( 3x^2 +11x -4) = -3x^2 -11x +4 (ii) -(3x-1)(2-x)(x-+4) 將上面做到個個代返入去... 即係-(3x-1)(x+4)= -3x^2 -11x +4 之後照乘 = ( - 3x^2 -11x +4 ) ( 2 - x ) =2( - 3x^2 -11x +4 ) - x( - 3x^2 -11x +4 ) = - 6x ^2 -22x +8 + 3x^3 -11x^2 -4x = 3x^3 -17x^2 -26x +8

其他解答:

(i) -(3x-1)(x+4) =(1-3x)(x+4) =1(x+4)-3x(x+4) =x+4-3x^2-12x =-3x^2-11x+4 (ii) -(3x-1)(2-x)(x+4) from(i) =(-3x^2-11x+4)(2-x) =-6x^2+3x^3-22x+11x^2+8-4x =3x^3+5x^2-26x+8 2007-10-03 17:30:26 補充： 下面兄弟，(-x)(-11x)=-11x^2嗎????