Whakaoti i te 3x^2+17x+10

$$3 { x }^{ 2 } +17x+10$$

Tauwehe

$$\left(x+5\right)\left(3x+2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3x^{2}+ax+bx+10. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=17 ab=3\times 10=30$$

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 30.

$$1,30 2,15 3,10 5,6$$

Tātaihia te tapeke mō ia takirua.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

Ko te otinga te takirua ka hoatu i te tapeke 17.

$$a=2 b=15$$

Tuhia anō te 3x^{2}+17x+10 hei \left(3x^{2}+2x\right)+\left(15x+10\right).

$$\left(3x^{2}+2x\right)+\left(15x+10\right)$$

Tauwehea te x i te tuatahi me te 5 i te rōpū tuarua.

$$x\left(3x+2\right)+5\left(3x+2\right)$$

Whakatauwehea atu te kīanga pātahi 3x+2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x+2\right)\left(x+5\right)$$

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.

$$3x^{2}+17x+10=0$$

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.

$$x=\frac{-17±\sqrt{17^{2}-4\times 3\times 10}}{2\times 3}$$

Pūrua 17.

$$x=\frac{-17±\sqrt{289-4\times 3\times 10}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-17±\sqrt{289-12\times 10}}{2\times 3}$$

Whakareatia -12 ki te 10.

$$x=\frac{-17±\sqrt{289-120}}{2\times 3}$$

Tāpiri 289 ki te -120.

$$x=\frac{-17±\sqrt{169}}{2\times 3}$$

Tuhia te pūtakerua o te 169.

$$x=\frac{-17±13}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-17±13}{6}$$

Nā, me whakaoti te whārite x=\frac{-17±13}{6} ina he tāpiri te ±. Tāpiri -17 ki te 13.

$$x=-\frac{4}{6}$$

Whakahekea te hautanga \frac{-4}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{2}{3}$$

Nā, me whakaoti te whārite x=\frac{-17±13}{6} ina he tango te ±. Tango 13 mai i -17.

$$x=-\frac{30}{6}$$

Whakawehe -30 ki te 6.

$$x=-5$$

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

$$3x^{2}+17x+10=3\left(x-\left(-\frac{2}{3}\right)\right)\left(x-\left(-5\right)\right)$$

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

$$3x^{2}+17x+10=3\left(x+\frac{2}{3}\right)\left(x+5\right)$$

Tāpiri \frac{2}{3} 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.

$$3x^{2}+17x+10=3\times \frac{3x+2}{3}\left(x+5\right)$$

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

$$3x^{2}+17x+10=\left(3x+2\right)\left(x+5\right)$$

Aromātai

$$\left(x+5\right)\left(3x+2\right)$$