Whakaoti i te 4x^2+10x+6

$$4 x ^ { 2 } + 10 x + 6$$

Tauwehe

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

Ngā Upane Otinga

Tauwehea te 2.

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

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

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

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

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

Tauwehea te 2x i te tuatahi me te 3 i te rōpū tuarua.

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

Whakatauwehea atu te kīanga pātahi x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

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

Me tuhi anō te kīanga whakatauwehe katoa.

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

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.

$$4x^{2}+10x+6=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 4\times 6}}{2\times 4}$$

Pūrua 10.

$$x=\frac{-10±\sqrt{100-4\times 4\times 6}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-10±\sqrt{100-16\times 6}}{2\times 4}$$

Whakareatia -16 ki te 6.

$$x=\frac{-10±\sqrt{100-96}}{2\times 4}$$

Tāpiri 100 ki te -96.

$$x=\frac{-10±\sqrt{4}}{2\times 4}$$

Tuhia te pūtakerua o te 4.

$$x=\frac{-10±2}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-10±2}{8}$$

Nā, me whakaoti te whārite x=\frac{-10±2}{8} ina he tāpiri te ±. Tāpiri -10 ki te 2.

$$x=-\frac{8}{8}$$

Whakawehe -8 ki te 8.

$$x=-1$$

Nā, me whakaoti te whārite x=\frac{-10±2}{8} ina he tango te ±. Tango 2 mai i -10.

$$x=-\frac{12}{8}$$

Whakahekea te hautanga \frac{-12}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

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

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

$$4x^{2}+10x+6=4\left(x-\left(-1\right)\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$4x^{2}+10x+6=4\left(x+1\right)\left(x+\frac{3}{2}\right)$$

Tāpiri \frac{3}{2} 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.

$$4x^{2}+10x+6=4\left(x+1\right)\times \frac{2x+3}{2}$$

Whakakorea atu te tauwehe pūnoa nui rawa 2 i roto i te 4 me te 2.

$$4x^{2}+10x+6=2\left(x+1\right)\left(2x+3\right)$$

Aromātai

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