Whakaoti i te f(x)=5x^2+10x-15

$$f ( x ) = 5 x ^ { 2 } + 10 x - 15$$

Tauwehe

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

Ngā Upane Otinga

Tauwehea te 5.

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

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

$$a+b=2 ab=1\left(-3\right)=-3$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-1 b=3$$

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

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

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

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

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

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

Me tuhi anō te kīanga whakatauwehe katoa.

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

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.

$$5x^{2}+10x-15=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 5\left(-15\right)}}{2\times 5}$$

Pūrua 10.

$$x=\frac{-10±\sqrt{100-4\times 5\left(-15\right)}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-10±\sqrt{100-20\left(-15\right)}}{2\times 5}$$

Whakareatia -20 ki te -15.

$$x=\frac{-10±\sqrt{100+300}}{2\times 5}$$

Tāpiri 100 ki te 300.

$$x=\frac{-10±\sqrt{400}}{2\times 5}$$

Tuhia te pūtakerua o te 400.

$$x=\frac{-10±20}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-10±20}{10}$$

Nā, me whakaoti te whārite x=\frac{-10±20}{10} ina he tāpiri te ±. Tāpiri -10 ki te 20.

$$x=\frac{10}{10}$$

Whakawehe 10 ki te 10.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{-10±20}{10} ina he tango te ±. Tango 20 mai i -10.

$$x=-\frac{30}{10}$$

Whakawehe -30 ki te 10.

$$x=-3$$

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 -3 mō te x_{2}.

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

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

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

Aromātai

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