Whakaoti i te 15x^2+25x+10

$$15 x ^ { 2 } + 25 x + 10$$

Tauwehe

$$5\left(x+1\right)\left(3x+2\right)$$

Ngā Upane Otinga

Tauwehea te 5.

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

Whakaarohia te 3x^{2}+5x+2. Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3x^{2}+ax+bx+2. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=5 ab=3\times 2=6$$

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 6.

$$1,6 2,3$$

Tātaihia te tapeke mō ia takirua.

$$1+6=7 2+3=5$$

Ko te otinga te takirua ka hoatu i te tapeke 5.

$$a=2 b=3$$

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

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

Whakatauwehea atu x i te 3x^{2}+2x.

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

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

$$\left(3x+2\right)\left(x+1\right)$$

Me tuhi anō te kīanga whakatauwehe katoa.

$$5\left(3x+2\right)\left(x+1\right)$$

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.

$$15x^{2}+25x+10=0$$

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.

$$x=\frac{-25±\sqrt{25^{2}-4\times 15\times 10}}{2\times 15}$$

Pūrua 25.

$$x=\frac{-25±\sqrt{625-4\times 15\times 10}}{2\times 15}$$

Whakareatia -4 ki te 15.

$$x=\frac{-25±\sqrt{625-60\times 10}}{2\times 15}$$

Whakareatia -60 ki te 10.

$$x=\frac{-25±\sqrt{625-600}}{2\times 15}$$

Tāpiri 625 ki te -600.

$$x=\frac{-25±\sqrt{25}}{2\times 15}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-25±5}{2\times 15}$$

Whakareatia 2 ki te 15.

$$x=\frac{-25±5}{30}$$

Nā, me whakaoti te whārite x=\frac{-25±5}{30} ina he tāpiri te ±. Tāpiri -25 ki te 5.

$$x=-\frac{20}{30}$$

Whakahekea te hautanga \frac{-20}{30} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

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

Nā, me whakaoti te whārite x=\frac{-25±5}{30} ina he tango te ±. Tango 5 mai i -25.

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

Whakawehe -30 ki te 30.

$$x=-1$$

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 -1 mō te x_{2}.

$$15x^{2}+25x+10=15\left(x-\left(-\frac{2}{3}\right)\right)\left(x-\left(-1\right)\right)$$

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

$$15x^{2}+25x+10=15\left(x+\frac{2}{3}\right)\left(x+1\right)$$

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.

$$15x^{2}+25x+10=15\times \frac{3x+2}{3}\left(x+1\right)$$

Whakakorea atu te tauwehe pūnoa nui rawa 3 i roto i te 15 me te 3.

$$15x^{2}+25x+10=5\left(3x+2\right)\left(x+1\right)$$

Aromātai

$$5\left(x+1\right)\left(3x+2\right)$$