Whakaoti i te y^3-2y^2+y-2=0

$$y ^ { 3 } - 2 y ^ { 2 } + y - 2 = 0$$

Whakaoti mō y

$$y=2$$

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 -2, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±2,±1$$

Kimihia tētahi pūtake pērā mā te whakamātau i ngā uara tau tōpū katoa, e tīmata ana i te mea iti rawa mā te uara pū. Mēnā kāore he pūtake tau tōpū e kitea, whakamātauria ngā hautanga.

$$y=2$$

Mā te whakatakotoranga Tauwehe, he tauwehe te y-k o te pūrau mō ia pūtake k. Whakawehea te y^{3}-2y^{2}+y-2 ki te y-2, kia riro ko y^{2}+1. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$y^{2}+1=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 1 mō te a, te 0 mō te b, me te 1 mō te c i te ture pūrua.

$$y=\frac{0±\sqrt{0^{2}-4\times 1\times 1}}{2}$$

Mahia ngā tātaitai.

$$y=\frac{0±\sqrt{-4}}{2}$$

Tā te mea e kore te pūrua o tētahi tau tōraro e tautohutia ki te āpure tūturu, kāhore he rongoā.

$$y\in \emptyset$$

Rārangitia ngā otinga katoa i kitea.

$$y=2$$

Whakaoti mō y (complex solution)

$$y=-i$$

$$y=2$$

$$y=i$$