Whakaoti i te 16x^2+8x

$$16 x ^ { 2 } + 8 x$$

Tauwehe

$$8x\left(2x+1\right)$$

Ngā Upane Otinga

Tauwehea te 8.

$$8\left(2x^{2}+x\right)$$

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

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

Me tuhi anō te kīanga whakatauwehe katoa.

$$8x\left(2x+1\right)$$

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.

$$16x^{2}+8x=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}}}{2\times 16}$$

Tuhia te pūtakerua o te 8^{2}.

$$x=\frac{-8±8}{2\times 16}$$

Whakareatia 2 ki te 16.

$$x=\frac{-8±8}{32}$$

Nā, me whakaoti te whārite x=\frac{-8±8}{32} ina he tāpiri te ±. Tāpiri -8 ki te 8.

$$x=\frac{0}{32}$$

Whakawehe 0 ki te 32.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-8±8}{32} ina he tango te ±. Tango 8 mai i -8.

$$x=-\frac{16}{32}$$

Whakahekea te hautanga \frac{-16}{32} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 16.

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

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 -\frac{1}{2} mō te x_{2}.

$$16x^{2}+8x=16x\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$16x^{2}+8x=16x\left(x+\frac{1}{2}\right)$$

Tāpiri \frac{1}{2} 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.

$$16x^{2}+8x=16x\times \frac{2x+1}{2}$$

Whakakorea atu te tauwehe pūnoa nui rawa 2 i roto i te 16 me te 2.

$$16x^{2}+8x=8x\left(2x+1\right)$$

Aromātai

$$8x\left(2x+1\right)$$