Whakaoti i te 3x^4+x^3+2x^2+4x-40

$$3 { x }^{ 4 } + { x }^{ 3 } +2 { x }^{ 2 } +4x-40$$

Tauwehe

$$\left(3x-5\right)\left(x+2\right)\left(x^{2}+4\right)$$

Ngā Upane Otinga

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

$$3x^{4}+x^{3}+2x^{2}+4x-40=0$$

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 -40, ā, ka wehea e q te whakarea arahanga 3. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{40}{3},±40,±\frac{20}{3},±20,±\frac{10}{3},±10,±\frac{8}{3},±8,±\frac{5}{3},±5,±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

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.

$$x=-2$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 3x^{4}+x^{3}+2x^{2}+4x-40 ki te x+2, kia riro ko 3x^{3}-5x^{2}+12x-20. Kia tauwehea ai te otinga, me whakaoti te whārite ina ōrite ki te 0.

$$3x^{3}-5x^{2}+12x-20=0$$

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 -20, ā, ka wehea e q te whakarea arahanga 3. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{20}{3},±20,±\frac{10}{3},±10,±\frac{5}{3},±5,±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

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.

$$x=\frac{5}{3}$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 3x^{3}-5x^{2}+12x-20 ki te 3\left(x-\frac{5}{3}\right)=3x-5, kia riro ko x^{2}+4. Kia tauwehea ai te otinga, me whakaoti te whārite ina ōrite ki te 0.

$$x^{2}+4=0$$

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 4 mō te c i te ture pūrua.

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

Mahia ngā tātaitai.

$$x=\frac{0±\sqrt{-16}}{2}$$

Kāore te pūrau x^{2}+4 i whakatauwehea i te mea kāhore ōna pūtake whakahau.

$$x^{2}+4$$

Me tuhi anō te kīanga whakatauwehe mā ngā pūtake i riro.

$$\left(3x-5\right)\left(x+2\right)\left(x^{2}+4\right)$$

Aromātai

$$\left(3x-5\right)\left(x+2\right)\left(x^{2}+4\right)$$