1題因式分解

1a)
b^2 - 10b + 21
我地可以先列哂b^2 term同constant term 既factor出黎先
b^2 得1x1, 21就有1x21, 3x7, -1x-21, -3x-7
搵一組roots加埋出到題目想要既"-10"就係答案
要留意題目要既係-10, 所以係-3 + (-7) = -10
所以b^2 - 10b + 21 = (b-7)(b-3)

1b)
(m+1)^2 - 10(m+1) + 21
明顯這題同a part係好類似的, 只不過係b變左做m+1
所以只要將b sub做(m+1)
就出到 [(m+1]-7][(m+1)-3] = (m-6)(m-2)



題外話：如果預到例如 2b^2 - 13b + 21
可以先將頭尾各自既factors寫哂出黎
2可以係1x2或者-1x-2
21就係 1x21, 3x7, -1x-21, -3x-7
用交叉相乘法[即左上乘右下 + 左下乘右上]
1 -1
X 出唔到題目想要既-13, 可以試第2組roots
2 -21

今次用
1 -3
X 2x(-3) + 1x(-7) = -13, 做到題目想要既答案喇
2 -7
所以2b^2 - 13b + 21 = (b-3)(2b-7){打橫咁抄出黎, 變左[(1)b + (-3)][(2)b + (-7)] }

1(a)
b^2-10b+21


b +3 +1 -3 -1
b +7 +21 -7 -21
__________________

3b+7b=10b
This doesn't give -10b.b+21b=22b
This doesn't give -10b.(-3)+(-7)=-10
This gives -10b!! ^^

So,
b^2-10b+21
=(b-3)(b-7)

(b)
(m+1)^2-10(m+1)+21


(m+1) +3 +1 -3 -1
(m+1) +7 +21 -7 -21
________________________

3(m+1)+7(m+1)=10(m+1)
This doesn't give -10(m+1).

1(m+1)+21(m+1)=22(m+1)
This doesn't give -10(m+1).

[-3(m+1)]+[-7(m+1)]=-10(m+1)
This gives -10(m+1)!!^^

So,
(m+1)^2-10(m+1)+21
=[(m+1)-3][(m+1)-7]
=(m+1-3)(m+1-7)
=(m-2)(m-6)


2010-10-03 23:07:31 補充：
The following website is the same question as yours! ^^

b^2-10b+21(b-3)(b-7)....................by Cross-method
____________________
(m+1)^2-10(m+1)+21

Let m+1 be b:
(m+1-3)(m+1-7)
(m-2)(m-6)

如上面的答案
其實第1題可以用計數機
你上網search下

1(a)
b^2-10b+21
=(b-7)(b-3)
1(b)
(m+1)^2-10(m+1)+21
=(m+1-7)(m+1-3)
=(m-6)(m-2)