Whakaoti i te 10x^2-35x+30

$$10 x ^ { 2 } - 35 x + 30$$

Tauwehe

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

Ngā Upane Otinga

Tauwehea te 5.

$$5\left(2x^{2}-7x+6\right)$$

Whakaarohia te 2x^{2}-7x+6. Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 2x^{2}+ax+bx+6. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-7 ab=2\times 6=12$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 12.

$$-1,-12 -2,-6 -3,-4$$

Tātaihia te tapeke mō ia takirua.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

Ko te otinga te takirua ka hoatu i te tapeke -7.

$$a=-4 b=-3$$

Tuhia anō te 2x^{2}-7x+6 hei \left(2x^{2}-4x\right)+\left(-3x+6\right).

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

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

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

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

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

Me tuhi anō te kīanga whakatauwehe katoa.

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

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.

$$10x^{2}-35x+30=0$$

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.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 10\times 30}}{2\times 10}$$

Pūrua -35.

$$x=\frac{-\left(-35\right)±\sqrt{1225-4\times 10\times 30}}{2\times 10}$$

Whakareatia -4 ki te 10.

$$x=\frac{-\left(-35\right)±\sqrt{1225-40\times 30}}{2\times 10}$$

Whakareatia -40 ki te 30.

$$x=\frac{-\left(-35\right)±\sqrt{1225-1200}}{2\times 10}$$

Tāpiri 1225 ki te -1200.

$$x=\frac{-\left(-35\right)±\sqrt{25}}{2\times 10}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-\left(-35\right)±5}{2\times 10}$$

Ko te tauaro o -35 ko 35.

$$x=\frac{35±5}{2\times 10}$$

Whakareatia 2 ki te 10.

$$x=\frac{35±5}{20}$$

Nā, me whakaoti te whārite x=\frac{35±5}{20} ina he tāpiri te ±. Tāpiri 35 ki te 5.

$$x=\frac{40}{20}$$

Whakawehe 40 ki te 20.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{35±5}{20} ina he tango te ±. Tango 5 mai i 35.

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

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

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

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

$$10x^{2}-35x+30=10\left(x-2\right)\left(x-\frac{3}{2}\right)$$

Tango \frac{3}{2} mai i x mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$10x^{2}-35x+30=10\left(x-2\right)\times \frac{2x-3}{2}$$

Whakakorea atu te tauwehe pūnoa nui rawa 2 i roto i te 10 me te 2.

$$10x^{2}-35x+30=5\left(x-2\right)\left(2x-3\right)$$

Aromātai

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