Whakaoti i te 3v^2+11v+8

$$3 v ^ { 2 } + 11 v + 8$$

Tauwehe

$$\left(v+1\right)\left(3v+8\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3v^{2}+av+bv+8. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=11 ab=3\times 8=24$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 24.

$$1,24 2,12 3,8 4,6$$

Tātaihia te tapeke mō ia takirua.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

Ko te otinga te takirua ka hoatu i te tapeke 11.

$$a=3 b=8$$

Tuhia anō te 3v^{2}+11v+8 hei \left(3v^{2}+3v\right)+\left(8v+8\right).

$$\left(3v^{2}+3v\right)+\left(8v+8\right)$$

Tauwehea te 3v i te tuatahi me te 8 i te rōpū tuarua.

$$3v\left(v+1\right)+8\left(v+1\right)$$

Whakatauwehea atu te kīanga pātahi v+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(v+1\right)\left(3v+8\right)$$

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.

$$3v^{2}+11v+8=0$$

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.

$$v=\frac{-11±\sqrt{11^{2}-4\times 3\times 8}}{2\times 3}$$

Pūrua 11.

$$v=\frac{-11±\sqrt{121-4\times 3\times 8}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$v=\frac{-11±\sqrt{121-12\times 8}}{2\times 3}$$

Whakareatia -12 ki te 8.

$$v=\frac{-11±\sqrt{121-96}}{2\times 3}$$

Tāpiri 121 ki te -96.

$$v=\frac{-11±\sqrt{25}}{2\times 3}$$

Tuhia te pūtakerua o te 25.

$$v=\frac{-11±5}{2\times 3}$$

Whakareatia 2 ki te 3.

$$v=\frac{-11±5}{6}$$

Nā, me whakaoti te whārite v=\frac{-11±5}{6} ina he tāpiri te ±. Tāpiri -11 ki te 5.

$$v=-\frac{6}{6}$$

Whakawehe -6 ki te 6.

$$v=-1$$

Nā, me whakaoti te whārite v=\frac{-11±5}{6} ina he tango te ±. Tango 5 mai i -11.

$$v=-\frac{16}{6}$$

Whakahekea te hautanga \frac{-16}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$v=-\frac{8}{3}$$

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 -1 mō te x_{1} me te -\frac{8}{3} mō te x_{2}.

$$3v^{2}+11v+8=3\left(v-\left(-1\right)\right)\left(v-\left(-\frac{8}{3}\right)\right)$$

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

$$3v^{2}+11v+8=3\left(v+1\right)\left(v+\frac{8}{3}\right)$$

Tāpiri \frac{8}{3} ki te v 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.

$$3v^{2}+11v+8=3\left(v+1\right)\times \frac{3v+8}{3}$$

Whakakorea atu te tauwehe pūnoa nui rawa 3 i roto i te 3 me te 3.

$$3v^{2}+11v+8=\left(v+1\right)\left(3v+8\right)$$

Aromātai

$$\left(v+1\right)\left(3v+8\right)$$