Whakaoti i te -5y^2-8y+4

$$- 5 y ^ { 2 } - 8 y + 4$$

Tauwehe

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

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei -5y^{2}+ay+by+4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-8 ab=-5\times 4=-20$$

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 -20.

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

Tātaihia te tapeke mō ia takirua.

$$1-20=-19 2-10=-8 4-5=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -8.

$$a=2 b=-10$$

Tuhia anō te -5y^{2}-8y+4 hei \left(-5y^{2}+2y\right)+\left(-10y+4\right).

$$\left(-5y^{2}+2y\right)+\left(-10y+4\right)$$

Tauwehea te -y i te tuatahi me te -2 i te rōpū tuarua.

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

Whakatauwehea atu te kīanga pātahi 5y-2 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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.

$$-5y^{2}-8y+4=0$$

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.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-5\right)\times 4}}{2\left(-5\right)}$$

Pūrua -8.

$$y=\frac{-\left(-8\right)±\sqrt{64-4\left(-5\right)\times 4}}{2\left(-5\right)}$$

Whakareatia -4 ki te -5.

$$y=\frac{-\left(-8\right)±\sqrt{64+20\times 4}}{2\left(-5\right)}$$

Whakareatia 20 ki te 4.

$$y=\frac{-\left(-8\right)±\sqrt{64+80}}{2\left(-5\right)}$$

Tāpiri 64 ki te 80.

$$y=\frac{-\left(-8\right)±\sqrt{144}}{2\left(-5\right)}$$

Tuhia te pūtakerua o te 144.

$$y=\frac{-\left(-8\right)±12}{2\left(-5\right)}$$

Ko te tauaro o -8 ko 8.

$$y=\frac{8±12}{2\left(-5\right)}$$

Whakareatia 2 ki te -5.

$$y=\frac{8±12}{-10}$$

Nā, me whakaoti te whārite y=\frac{8±12}{-10} ina he tāpiri te ±. Tāpiri 8 ki te 12.

$$y=\frac{20}{-10}$$

Whakawehe 20 ki te -10.

$$y=-2$$

Nā, me whakaoti te whārite y=\frac{8±12}{-10} ina he tango te ±. Tango 12 mai i 8.

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

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

$$y=\frac{2}{5}$$

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

$$-5y^{2}-8y+4=-5\left(y-\left(-2\right)\right)\left(y-\frac{2}{5}\right)$$

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

$$-5y^{2}-8y+4=-5\left(y+2\right)\left(y-\frac{2}{5}\right)$$

Tango \frac{2}{5} mai i y mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$-5y^{2}-8y+4=-5\left(y+2\right)\times \frac{-5y+2}{-5}$$

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

$$-5y^{2}-8y+4=\left(y+2\right)\left(-5y+2\right)$$

Aromātai

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