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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2007-10-03 2:41 am
回答 (2)
2007-10-03 2:51 am
✔ 最佳答案
(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
2007-10-03 2:49 am
(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????
-(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????
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