Whakaoti i te f(x)=x^3-6x^2+11x-6.g(x)=x+2

$$f ( x ) = x ^ { 3 } - 6 x ^ { 2 } + 11 x - 6 . g ( x ) = x + 2$$

Tauwehe

$$\left(x-3\right)\left(x-2\right)\left(x-1\right)$$

Ngā Upane Otinga

Tā te Rational Root Theorem, ko ngā pūtake whakahau katoa o tētahi pūrau kei te āhua o \frac{p}{q}, ina wehea e p te kīanga pūmau -6, ā, ka wehea e q te whakarea arahanga 1. Ko tetahi pūtake pērā ko 3. Tauwehea te pūrau mā te whakawehe mā te x-3.

$$\left(x-3\right)\left(x^{2}-3x+2\right)$$

Whakaarohia te x^{2}-3x+2. Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx+2. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-3 ab=1\times 2=2$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-2 b=-1$$

Tuhia anō te x^{2}-3x+2 hei \left(x^{2}-2x\right)+\left(-x+2\right).

$$\left(x^{2}-2x\right)+\left(-x+2\right)$$

Tauwehea te x i te tuatahi me te -1 i te rōpū tuarua.

$$x\left(x-2\right)-\left(x-2\right)$$

Whakatauwehea atu te kīanga pātahi x-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-2\right)\left(x-1\right)$$

Me tuhi anō te kīanga whakatauwehe katoa.

$$\left(x-3\right)\left(x-2\right)\left(x-1\right)$$

Aromātai

$$\left(x-3\right)\left(x-2\right)\left(x-1\right)$$