Whakaoti i te 12x^2+80x+100

$$12 { x }^{ 2 } +80x+100$$

Tauwehe

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

Ngā Upane Otinga

Tauwehea te 4.

$$4\left(3x^{2}+20x+25\right)$$

Whakaarohia te 3x^{2}+20x+25. Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3x^{2}+ax+bx+25. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=20 ab=3\times 25=75$$

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 75.

$$1,75 3,25 5,15$$

Tātaihia te tapeke mō ia takirua.

$$1+75=76 3+25=28 5+15=20$$

Ko te otinga te takirua ka hoatu i te tapeke 20.

$$a=5 b=15$$

Tuhia anō te 3x^{2}+20x+25 hei \left(3x^{2}+5x\right)+\left(15x+25\right).

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

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

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

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

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

Me tuhi anō te kīanga whakatauwehe katoa.

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

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.

$$12x^{2}+80x+100=0$$

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.

$$x=\frac{-80±\sqrt{80^{2}-4\times 12\times 100}}{2\times 12}$$

Pūrua 80.

$$x=\frac{-80±\sqrt{6400-4\times 12\times 100}}{2\times 12}$$

Whakareatia -4 ki te 12.

$$x=\frac{-80±\sqrt{6400-48\times 100}}{2\times 12}$$

Whakareatia -48 ki te 100.

$$x=\frac{-80±\sqrt{6400-4800}}{2\times 12}$$

Tāpiri 6400 ki te -4800.

$$x=\frac{-80±\sqrt{1600}}{2\times 12}$$

Tuhia te pūtakerua o te 1600.

$$x=\frac{-80±40}{2\times 12}$$

Whakareatia 2 ki te 12.

$$x=\frac{-80±40}{24}$$

Nā, me whakaoti te whārite x=\frac{-80±40}{24} ina he tāpiri te ±. Tāpiri -80 ki te 40.

$$x=-\frac{40}{24}$$

Whakahekea te hautanga \frac{-40}{24} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

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

Nā, me whakaoti te whārite x=\frac{-80±40}{24} ina he tango te ±. Tango 40 mai i -80.

$$x=-\frac{120}{24}$$

Whakawehe -120 ki te 24.

$$x=-5$$

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

$$12x^{2}+80x+100=12\left(x-\left(-\frac{5}{3}\right)\right)\left(x-\left(-5\right)\right)$$

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

$$12x^{2}+80x+100=12\left(x+\frac{5}{3}\right)\left(x+5\right)$$

Tāpiri \frac{5}{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.

$$12x^{2}+80x+100=12\times \frac{3x+5}{3}\left(x+5\right)$$

Whakakorea atu te tauwehe pūnoa nui rawa 3 i roto i te 12 me te 3.

$$12x^{2}+80x+100=4\left(3x+5\right)\left(x+5\right)$$

Aromātai

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