F.3 MATHS! FACTORIZATION!!!!!

2007-12-22 5:34 am
FACTARIZE THE FOLLOWING:
10x^2-12x^3+8x

16a^2-15b^2+8ab

(x+2)^2-4(x^2-4x-12)

27s^3-125t^3-6s+10t

CLEAR STEPS PLZ~ =]

回答 (2)

2007-12-22 6:36 am
頭2題係Cross method,冇乜Step可言,以下係我一點心得:
假設要Factorize嘅係Ax^2+Bx+C 或者係 Ax^2-Bx+C.(Ax^2和C同為正號)
計AxC,假設係D
找出兩個整數E和F,使得ExF=D而且E+F=B
然後Ax^2+Bx+C = A^2+Ex+Fx+C(因為E+F=B)或者Ax^2-Bx+C = A^2-Ex-Fx+C
最後用Grouping terms就計到
例子:6x^2-11x+4
好清楚,A=6,B=11,C=4

D=AxC=6x4=24

24=1x24(1+24=25)
=2x12(2+12=14)
=3x8(3+8=11=B)
=4x6(4+6=10)
當E=3,F=8時,ExF=3x8=24=D且E+F=3+8=11=B
所以6x^2-11x+4=6x^2-3x-8x+4
=(6x^2-3x)-(8x-4)
=3x(2x-1)-4(2x-1)
=(2x-1)(3x-4)

但係如果要Factorize嘅係Ax^2+Bx-C 或者係 Ax^2-Bx-C.(Ax^2和C為一正一負)
計AxC,假設係D
找出兩個整數E和F,使得ExF=D而且E-F=B
然後Ax^2+Bx-C = A^2+Ex-Fx-C(因為E-F=B)或者Ax^2-Bx-C = A^2-Ex+Fx-C
最後用Grouping terms就計到

例子:
16a^2-15b^2+8ab=16a^2+8ab-15b^2
A=16,B=8,C=15
D=AxC=16x15=240

240=1x240(240-1=239)
=2x120(120-2=118)
=3x80(80-3=77)
=4x60(60-4=56)
=5x48(48-5=43)
=6x40(40-6=34)
=8x30(30-8=22)
=10x24(24-10=14)
=12x20(20-12=8=B)

16a^2+8ab-15b^2 = 16a^2+20ab-12ab-15b^2
=(16a^2+20ab)-(12ab+15b^2)
=4a(4a+5b)-3b(4a+5b)
=(4a+5b)(4a-3b)
--------------------------------------------------------------------------------------
10x^2-12x^3+8x=-2x(6x^2-5x-4)
A=6,B=5,C=4
D=AxC=6x4=24

24=1x24(24-1=23)
=2x12(12-2=10)
=3x8(8-3=5=B)

-2x(6x^2-5x-4)=-2x(6x^2-8x+3x-4)
=-2x[(6x^2+3x)-(8x+4)]
=-2x[3x(2x+1)-4(2x+1)]
=-2x(2x+1)(3x-4)
=2x(2x+1)(4-3x)

第3題係Cross method(或者直接Expand咗之後再用Factorize)
(x+2)^2-4(x^2-4x-12)
=(x+2)^2-4(x+2)(x-6)-------------------------Cross method,自己唸吓點計
=(x+2)[(x+2)-4(x-6)]
=(x+2)(26-3x)

第4題唔可以直接Group埋27s^3同-6s一組,一定要用Difference of 2 Cubes計(27s^3-125t^3)
27s^3-125t^3-6s+10t
=(27s^3-125t^3)-(6s-10t)
=(3s-5t)(9s^2+15st+25t^2)-2(3s-5t)----------------------------Difference of 2 Cubes
=(3s-5t)(9s^2+15st+25t^2-2)

Cross method 其實唔使寫Steps,所以頭1題應該咁寫:
10x^2-12x^3+8x
=-2x(6x^2-5x-4)
=-2x(2x+1)(3x-4)-----------------------中間部驟全部Skip
=2x(2x+1)(4-3x)
2007-12-22 5:56 am
10x^2-12x^3+8x
=-2x(6x^2-5x-4)
=-2x(3x-4)(2x+1)

16a^2-15b^2+8ab
=16a^2+8ab-15b^2
=(4a+5b)(4a-3b)

(x+2)^2-4(x^2-4x-12)
=(x+2)^2-4(x+2)(x-6)
=(x+2)[x+2-4(x-6)]
=(x+2)[-3x+26]
=-(x+2)(3x-26)

27s^3-125t^3-6s+10t
=(3s)^3-(5t)^3 - 2(3s-5t)
=(3s-5t)(9s^2+15st+25t^2) - 2(3s-5t)
=(3s-5t)[9s^2+15st+25t^2 - 2]
參考: me


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