Whakaoti i te 5x^2+23x+12

$$5 { x }^{ 2 } +23x+12$$

Tauwehe

$$\left(x+4\right)\left(5x+3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 5x^{2}+ax+bx+12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=23 ab=5\times 12=60$$

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 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Tātaihia te tapeke mō ia takirua.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

Ko te otinga te takirua ka hoatu i te tapeke 23.

$$a=3 b=20$$

Tuhia anō te 5x^{2}+23x+12 hei \left(5x^{2}+3x\right)+\left(20x+12\right).

$$\left(5x^{2}+3x\right)+\left(20x+12\right)$$

Tauwehea te x i te tuatahi me te 4 i te rōpū tuarua.

$$x\left(5x+3\right)+4\left(5x+3\right)$$

Whakatauwehea atu te kīanga pātahi 5x+3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(5x+3\right)\left(x+4\right)$$

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.

$$5x^{2}+23x+12=0$$

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.

$$x=\frac{-23±\sqrt{23^{2}-4\times 5\times 12}}{2\times 5}$$

Pūrua 23.

$$x=\frac{-23±\sqrt{529-4\times 5\times 12}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-23±\sqrt{529-20\times 12}}{2\times 5}$$

Whakareatia -20 ki te 12.

$$x=\frac{-23±\sqrt{529-240}}{2\times 5}$$

Tāpiri 529 ki te -240.

$$x=\frac{-23±\sqrt{289}}{2\times 5}$$

Tuhia te pūtakerua o te 289.

$$x=\frac{-23±17}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-23±17}{10}$$

Nā, me whakaoti te whārite x=\frac{-23±17}{10} ina he tāpiri te ±. Tāpiri -23 ki te 17.

$$x=-\frac{6}{10}$$

Whakahekea te hautanga \frac{-6}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{3}{5}$$

Nā, me whakaoti te whārite x=\frac{-23±17}{10} ina he tango te ±. Tango 17 mai i -23.

$$x=-\frac{40}{10}$$

Whakawehe -40 ki te 10.

$$x=-4$$

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 -\frac{3}{5} mō te x_{1} me te -4 mō te x_{2}.

$$5x^{2}+23x+12=5\left(x-\left(-\frac{3}{5}\right)\right)\left(x-\left(-4\right)\right)$$

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

$$5x^{2}+23x+12=5\left(x+\frac{3}{5}\right)\left(x+4\right)$$

Tāpiri \frac{3}{5} 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.

$$5x^{2}+23x+12=5\times \frac{5x+3}{5}\left(x+4\right)$$

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

$$5x^{2}+23x+12=\left(5x+3\right)\left(x+4\right)$$

Aromātai

$$\left(x+4\right)\left(5x+3\right)$$