F.2 Maths

As follows:


圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/13-15.jpg



圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/14-12.jpg


2010-07-17 18:53:20 補充：
http://i707.photobucket.com/albums/ww74/stevieg90/13-15.jpg
http://i707.photobucket.com/albums/ww74/stevieg90/14-12.jpg

For 12(d), the prime factors are 2,3 and 19.

Factorization of polynomials

10)(a) Factorize 1+m+n+mn
=(1+m)+n(1+m)
=(1+n)(1+m).
(b) Hence,factorize 1+m+n+p+mn+mp+np+mnp.
=p(1+m+n+mn)+(1+m+n+mn)
=(p+1)(1+m+n+mn)
11B)Factorize (x²+3x)²+3(x²+3x)+2.
=(x²+3x)(x²+3x+3)+2
=(x²+3x)(x²+3x+2)+x²+3x+2
=(x²+3x)(x+1)(x+2)+(x+1)(x+2)
=(x+1)(x+2)(x²+3x+1)

12c)Hence,factorize 2x^3 +7x²+4x-4
=2x^3+7x²-4x-4+8x
=(x²+4x)(2x-1)+4(2x-1)
=(2x-1)(x²+4x+4)
=(2x-1)(x+2)²

d)Using the result of (c),find all the prime factor of 2 736.

唔明



Rate and ratio

If 2x=3y=4z , find(x-y+z) : (x+y-z).
x:y=3:2
x:z=2:1
so , x:y:z=6:4:3
so,(6-4+3):(6+4-3)
=5:7

2010-07-17 18:59:00 補充：
the prime factors are 2,3 and 19.

10a) 1+m+n+mn
=1+m+n(1+m)
=(1+m)(1+n)
10b) 1+m+n+p+mn+mp+np+mnp
=1+m+n+mn+p(1+m+n+mn)
=(1+m+n+mn)(1+p)
=(1+m)(1+n)(1+p)

11b) (x²+3x)²+3(x²+3x)+2
=(x²+3x+1)(x²+3x+2)
= (x²+3x+1)(x+1)(x+2)

12where is a and b???

c) the answer is (x+2)(2x^2+3x-2)
=(x+2)^2(2x-1)
d) you should use (a) or (b) , try to substitute some numbers

ratio

from 2x=3y=4z, y=2/3x, z=1/2x

so x-y+z =5/6x ; x+y-z= 7/6x
(x-y+z) : (x+y-z) = 5:7