解 2x^3 - x^2 - 72x + 36 =0

2x^3 - x^2 - 72x + 36 =0
2x^3 - 72x - x^2 + 36 = 0 [將佢d 執下d 位]
2x(x^2-36)-(x^2-36) = 0 [之後抽個x 出來,後面又抽個- 出來]
(x^2-36)(2x-1)=0 [之後出現同類項(x^2-36)~then 抽埋佢出黎]
(x+6)(x-6)(2x-1)=0 [將個(x^2-36)~~化簡佢]
(x+6)=0 => x=-6
(x-6)=0 => x=6
(2x-1)=0 => x=1/2
所以 x=6,-6,1/2

設f(x)=2x^3 - x^2 - 72x + 36 =0
f(6)=2(6)^3-(6)^2-72(6)+36
f(6)=0
之後便是短除,需要過程才找我吧,
f(x)=(x-6)(2x^2+11x-6)
拿(2x^2+11x-6)去做因式分解,
f(x)=(x-6)(2x-1)(x+6)
所以2x^3 - x^2 - 72x + 36 =0
(x-6)(2x^2+11x-6)=0
(x-6)(2x-1)(x+6)=0
x=6或0.5或-6