Whakaoti i te -5x^2-2x

$$- 5 x ^ { 2 } - 2 x$$

Tauwehe

$$-x\left(5x+2\right)$$

Ngā Upane Otinga

Tauwehea te x.

$$x\left(-5x-2\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Ka taea te huamaha pūrua te tauwehe mā te whakamahi i te huringa ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), ina ko x_{1} me x_{2} ngā otinga o te whārite pūrua ax^{2}+bx+c=0.

$$-5x^{2}-2x=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}}}{2\left(-5\right)}$$

Tuhia te pūtakerua o te \left(-2\right)^{2}.

$$x=\frac{-\left(-2\right)±2}{2\left(-5\right)}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2}{2\left(-5\right)}$$

Whakareatia 2 ki te -5.

$$x=\frac{2±2}{-10}$$

Nā, me whakaoti te whārite x=\frac{2±2}{-10} ina he tāpiri te ±. Tāpiri 2 ki te 2.

$$x=\frac{4}{-10}$$

Whakahekea te hautanga \frac{4}{-10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{2}{5}$$

Nā, me whakaoti te whārite x=\frac{2±2}{-10} ina he tango te ±. Tango 2 mai i 2.

$$x=\frac{0}{-10}$$

Whakawehe 0 ki te -10.

$$x=0$$

Tauwehea te kīanga taketake mā te whakamahi i te ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Me whakakapi te -\frac{2}{5} mō te x_{1} me te 0 mō te x_{2}.

$$-5x^{2}-2x=-5\left(x-\left(-\frac{2}{5}\right)\right)x$$

Whakamāmātia ngā kīanga katoa o te āhua p-\left(-q\right) ki te p+q.

$$-5x^{2}-2x=-5\left(x+\frac{2}{5}\right)x$$

Tāpiri \frac{2}{5} ki te x mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$-5x^{2}-2x=-5\times \frac{-5x-2}{-5}x$$

Whakakorea atu te tauwehe pūnoa nui rawa 5 i roto i te -5 me te -5.

$$-5x^{2}-2x=\left(-5x-2\right)x$$

Aromātai

$$-x\left(5x+2\right)$$