Whakaoti i te f(x)=x^2-3x-4

$$f ( x ) = x ^ { 2 } - 3 x - 4$$

Tauwehe

$$\left(x-4\right)\left(x+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx-4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-3 ab=1\left(-4\right)=-4$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -4.

$$1,-4 2,-2$$

Tātaihia te tapeke mō ia takirua.

$$1-4=-3 2-2=0$$

Ko te otinga te takirua ka hoatu i te tapeke -3.

$$a=-4 b=1$$

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

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

Whakatauwehea atu x i te x^{2}-4x.

$$x\left(x-4\right)+x-4$$

Whakatauwehea atu te kīanga pātahi x-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-4\right)\left(x+1\right)$$

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.

$$x^{2}-3x-4=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-4\right)}}{2}$$

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-4\right)}}{2}$$

Whakareatia -4 ki te -4.

$$x=\frac{-\left(-3\right)±\sqrt{9+16}}{2}$$

Tāpiri 9 ki te 16.

$$x=\frac{-\left(-3\right)±\sqrt{25}}{2}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-\left(-3\right)±5}{2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±5}{2}$$

Nā, me whakaoti te whārite x=\frac{3±5}{2} ina he tāpiri te ±. Tāpiri 3 ki te 5.

$$x=\frac{8}{2}$$

Whakawehe 8 ki te 2.

$$x=4$$

Nā, me whakaoti te whārite x=\frac{3±5}{2} ina he tango te ±. Tango 5 mai i 3.

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

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 4 mō te x_{1} me te -1 mō te x_{2}.

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

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

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

Aromātai

$$\left(x-4\right)\left(x+1\right)$$