Whakaoti i te (2x^2+2x)=

$$( 2 x ^ { 2 } + 2 x ) =$$

Tauwehe

$$2x\left(x+1\right)$$

Ngā Upane Otinga

Tauwehea te 2.

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

Whakaarohia te x^{2}+x. Tauwehea te x.

$$x\left(x+1\right)$$

Me tuhi anō te kīanga whakatauwehe katoa.

$$2x\left(x+1\right)$$

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.

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

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.

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

Tuhia te pūtakerua o te 2^{2}.

$$x=\frac{-2±2}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-2±2}{4}$$

Nā, me whakaoti te whārite x=\frac{-2±2}{4} ina he tāpiri te ±. Tāpiri -2 ki te 2.

$$x=\frac{0}{4}$$

Whakawehe 0 ki te 4.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-2±2}{4} ina he tango te ±. Tango 2 mai i -2.

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

Whakawehe -4 ki te 4.

$$x=-1$$

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 0 mō te x_{1} me te -1 mō te x_{2}.

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

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

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

Aromātai

$$2x\left(x+1\right)$$