Whakaoti i te x^5+5x^4+7x^3-x^2-8x-4

$$x ^ { 5 } + 5 x ^ { 4 } + 7 x ^ { 3 } - x ^ { 2 } - 8 x - 4$$

Tauwehe

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

Ngā Upane Otinga

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

$$x^{5}+5x^{4}+7x^{3}-x^{2}-8x-4=0$$

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 -4, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±4,±2,±1$$

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.

$$x=1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{5}+5x^{4}+7x^{3}-x^{2}-8x-4 ki te x-1, kia riro ko x^{4}+6x^{3}+13x^{2}+12x+4. Kia tauwehea ai te otinga, me whakaoti te whārite ina ōrite ki te 0.

$$x^{4}+6x^{3}+13x^{2}+12x+4=0$$

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

$$±4,±2,±1$$

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.

$$x=-1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{4}+6x^{3}+13x^{2}+12x+4 ki te x+1, kia riro ko x^{3}+5x^{2}+8x+4. Kia tauwehea ai te otinga, me whakaoti te whārite ina ōrite ki te 0.

$$x^{3}+5x^{2}+8x+4=0$$

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

$$±4,±2,±1$$

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.

$$x=-1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{3}+5x^{2}+8x+4 ki te x+1, kia riro ko x^{2}+4x+4. Kia tauwehea ai te otinga, me whakaoti te whārite ina ōrite ki te 0.

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

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 4 mō te b, me te 4 mō te c i te ture pūrua.

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

Mahia ngā tātaitai.

$$x=\frac{-4±0}{2}$$

He ōrite ngā whakatau.

$$x=-2$$

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

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

Aromātai

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