Whakaoti i te F(x+4)=x^2-6x+5

$$F ( x + 4 ) = x ^ { 2 } - 6 x + 5$$

Tauwehe

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

Ngā Upane Otinga

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

$$a+b=-6 ab=1\times 5=5$$

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. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-5 b=-1$$

Tuhia anō te x^{2}-6x+5 hei \left(x^{2}-5x\right)+\left(-x+5\right).

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

Tauwehea te x i te tuatahi me te -1 i te rōpū tuarua.

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

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

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

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.

$$x^{2}-6x+5=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5}}{2}$$

Pūrua -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 5}}{2}$$

Whakareatia -4 ki te 5.

$$x=\frac{-\left(-6\right)±\sqrt{36-20}}{2}$$

Tāpiri 36 ki te -20.

$$x=\frac{-\left(-6\right)±\sqrt{16}}{2}$$

Tuhia te pūtakerua o te 16.

$$x=\frac{-\left(-6\right)±4}{2}$$

Ko te tauaro o -6 ko 6.

$$x=\frac{6±4}{2}$$

Nā, me whakaoti te whārite x=\frac{6±4}{2} ina he tāpiri te ±. Tāpiri 6 ki te 4.

$$x=\frac{10}{2}$$

Whakawehe 10 ki te 2.

$$x=5$$

Nā, me whakaoti te whārite x=\frac{6±4}{2} ina he tango te ±. Tango 4 mai i 6.

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

Whakawehe 2 ki te 2.

$$x=1$$

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 5 mō te x_{1} me te 1 mō te x_{2}.

$$x^{2}-6x+5=\left(x-5\right)\left(x-1\right)$$

Aromātai

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