mathes問題一問

12v-12v^2-6u+3u^2
=3u^2-12v^2+12v-6u
=3(u^2-4v^2)+6(2v-u)
=3(u-2v)(u+2v)+6(2v-u)
=3(u-2v)(u+2v)-6(u-2v) <----- 2v-u = -(u-2v)
=3(u-2v)(u+2v)-3(u-2v)(2)
=3(u-2v)(2v+u-2),, <----------第1項和第2項都有"3(u-2v)" ,抽common factor, 第1項得番u+2v, 第2項得番"-2"

Ans: 12v-12v^2-6u+3u^2 = (12v - 6u) - 12v^2 + 3u^2
= 6(2v - u) - 3(4v^2 - u^2)
= 6(2v - u) - 3(2v - u)(2v + u)
= 2乘3(2v - u) - 3(2v - u)(2v + u) -------------(1)
= 3(2v - u)(2 - (2v + u)) -------------(2)
= 3(2v - u)(2 - 2v - u) ------------- (3)
= 3(u - 2v)(2v +u - 2) ------------(4)

如何由(1)變成(2)? 6(2v - u) = 2 X 3(2v - u)
之後抽comment factor 3(2v - u),所以剩下個2
彧者你假設3(2v - u)是x,你會看出是怎麼回事：

6(2v - u) - 3(2v - u)(2v + u) = 2x - x(2v + u)
= x(2 - 2v - u)
= --------------(1)

如何由(3)變成(4)?你把(2v - u)抽過負1, 即是(-1)(u - 2v)(2 - 2v - u)
再將(-1)乘入(2 - 2v - u),即是(-1)(2 - 2v - u) = (2v + u - 2)
所以形成(2v + u - 2)(u - 2v) ----------(4)