Whakaoti i te 2x^4+3x^3-8x^2-9x+6

$$2 x ^ { 4 } + 3 x ^ { 3 } - 8 x ^ { 2 } - 9 x + 6$$

Tauwehe

$$\left(2x-1\right)\left(x+2\right)\left(x^{2}-3\right)$$

Ngā Upane Otinga

Kia tauwehea ai te kīanga, me whakaoti te whārite ina ōrite ki te 0.

$$2x^{4}+3x^{3}-8x^{2}-9x+6=0$$

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

$$±3,±6,±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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.

$$x=-2$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 2x^{4}+3x^{3}-8x^{2}-9x+6 ki te x+2, kia riro ko 2x^{3}-x^{2}-6x+3. Kia tauwehea ai te otinga, me whakaoti te whārite ina ōrite ki te 0.

$$2x^{3}-x^{2}-6x+3=0$$

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

$$±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$x=\frac{1}{2}$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 2x^{3}-x^{2}-6x+3 ki te 2\left(x-\frac{1}{2}\right)=2x-1, kia riro ko x^{2}-3. Kia tauwehea ai te otinga, me whakaoti te whārite ina ōrite ki te 0.

$$x^{2}-3=0$$

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 -3 mō te c i te ture pūrua.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-3\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{0±2\sqrt{3}}{2}$$

Whakaotia te whārite x^{2}-3=0 ina he tōrunga te ±, ina he tōraro te ±.

$$x=-\sqrt{3} x=\sqrt{3}$$

Me tuhi anō te kīanga whakatauwehe mā ngā pūtake i riro. Kāore te pūrau x^{2}-3 i whakatauwehea i te mea kāhore ōna pūtake whakahau.

$$\left(2x-1\right)\left(x+2\right)\left(x^{2}-3\right)$$

Aromātai

$$\left(2x-1\right)\left(x+2\right)\left(x^{2}-3\right)$$