續因式分解.....要有直式

a)
b^2-10b+21


b +3 +1 -3 -1
b +7 +21 -7 -21
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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
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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)

b^2-10b+21
b^2-7b-3b+21
b(b-7)-3(b-7)
(b-3)(b-7)
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Let b be (m+1):
b^2-10b+21
(b-7)(b-3)
(m+1-7)(m+1-3)
(m-6)(m-2)