Whakaoti i te 2n^2-3n-20

$$2 n ^ { 2 } - 3 n - 20$$

Tauwehe

$$\left(n-4\right)\left(2n+5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 2n^{2}+an+bn-20. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-3 ab=2\left(-20\right)=-40$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -40.

$$1,-40 2,-20 4,-10 5,-8$$

Tātaihia te tapeke mō ia takirua.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

Ko te otinga te takirua ka hoatu i te tapeke -3.

$$a=-8 b=5$$

Tuhia anō te 2n^{2}-3n-20 hei \left(2n^{2}-8n\right)+\left(5n-20\right).

$$\left(2n^{2}-8n\right)+\left(5n-20\right)$$

Tauwehea te 2n i te tuatahi me te 5 i te rōpū tuarua.

$$2n\left(n-4\right)+5\left(n-4\right)$$

Whakatauwehea atu te kīanga pātahi n-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(n-4\right)\left(2n+5\right)$$

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.

$$2n^{2}-3n-20=0$$

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.

$$n=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-20\right)}}{2\times 2}$$

Pūrua -3.

$$n=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-20\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$n=\frac{-\left(-3\right)±\sqrt{9-8\left(-20\right)}}{2\times 2}$$

Whakareatia -8 ki te -20.

$$n=\frac{-\left(-3\right)±\sqrt{9+160}}{2\times 2}$$

Tāpiri 9 ki te 160.

$$n=\frac{-\left(-3\right)±\sqrt{169}}{2\times 2}$$

Tuhia te pūtakerua o te 169.

$$n=\frac{-\left(-3\right)±13}{2\times 2}$$

Ko te tauaro o -3 ko 3.

$$n=\frac{3±13}{2\times 2}$$

Whakareatia 2 ki te 2.

$$n=\frac{3±13}{4}$$

Nā, me whakaoti te whārite n=\frac{3±13}{4} ina he tāpiri te ±. Tāpiri 3 ki te 13.

$$n=\frac{16}{4}$$

Whakawehe 16 ki te 4.

$$n=4$$

Nā, me whakaoti te whārite n=\frac{3±13}{4} ina he tango te ±. Tango 13 mai i 3.

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

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

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

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 4 mō te x_{1} me te -\frac{5}{2} mō te x_{2}.

$$2n^{2}-3n-20=2\left(n-4\right)\left(n-\left(-\frac{5}{2}\right)\right)$$

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

$$2n^{2}-3n-20=2\left(n-4\right)\left(n+\frac{5}{2}\right)$$

Tāpiri \frac{5}{2} ki te n 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.

$$2n^{2}-3n-20=2\left(n-4\right)\times \frac{2n+5}{2}$$

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

$$2n^{2}-3n-20=\left(n-4\right)\left(2n+5\right)$$

Aromātai

$$\left(n-4\right)\left(2n+5\right)$$