Whakaoti i te (x-3)(x-25)(x+1)=03

$$(x-3)(x-25)(x+1)=03$$

Whakaoti mō x

$$x=25$$

$$x=3$$

$$x=-1$$

Ngā Upane Otinga

Whakamahia te āhuatanga tuaritanga hei whakarea te x-3 ki te x-25 ka whakakotahi i ngā kupu rite.

$$\left(x^{2}-28x+75\right)\left(x+1\right)=0\times 3$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x^{2}-28x+75 ki te x+1 ka whakakotahi i ngā kupu rite.

$$x^{3}-27x^{2}+47x+75=0\times 3$$

Whakareatia te 0 ki te 3, ka 0.

$$x^{3}-27x^{2}+47x+75=0$$

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

$$±75,±25,±15,±5,±3,±1$$

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.

$$x=-1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{3}-27x^{2}+47x+75 ki te x+1, kia riro ko x^{2}-28x+75. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$x^{2}-28x+75=0$$

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 -28 mō te b, me te 75 mō te c i te ture pūrua.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 1\times 75}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{28±22}{2}$$

Whakaotia te whārite x^{2}-28x+75=0 ina he tōrunga te ±, ina he tōraro te ±.

$$x=3 x=25$$

Rārangitia ngā otinga katoa i kitea.

$$x=-1 x=3 x=25$$