Whakaoti i te x^2+10x-56?

$$x ^ { 2 } + 10 x - 56 ?$$

Tauwehe

$$\left(x-4\right)\left(x+14\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx-56. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=10 ab=1\left(-56\right)=-56$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -56.

$$-1,56 -2,28 -4,14 -7,8$$

Tātaihia te tapeke mō ia takirua.

$$-1+56=55 -2+28=26 -4+14=10 -7+8=1$$

Ko te otinga te takirua ka hoatu i te tapeke 10.

$$a=-4 b=14$$

Tuhia anō te x^{2}+10x-56 hei \left(x^{2}-4x\right)+\left(14x-56\right).

$$\left(x^{2}-4x\right)+\left(14x-56\right)$$

Tauwehea te x i te tuatahi me te 14 i te rōpū tuarua.

$$x\left(x-4\right)+14\left(x-4\right)$$

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

$$\left(x-4\right)\left(x+14\right)$$

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.

$$x^{2}+10x-56=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-56\right)}}{2}$$

Pūrua 10.

$$x=\frac{-10±\sqrt{100-4\left(-56\right)}}{2}$$

Whakareatia -4 ki te -56.

$$x=\frac{-10±\sqrt{100+224}}{2}$$

Tāpiri 100 ki te 224.

$$x=\frac{-10±\sqrt{324}}{2}$$

Tuhia te pūtakerua o te 324.

$$x=\frac{-10±18}{2}$$

Nā, me whakaoti te whārite x=\frac{-10±18}{2} ina he tāpiri te ±. Tāpiri -10 ki te 18.

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

Whakawehe 8 ki te 2.

$$x=4$$

Nā, me whakaoti te whārite x=\frac{-10±18}{2} ina he tango te ±. Tango 18 mai i -10.

$$x=-\frac{28}{2}$$

Whakawehe -28 ki te 2.

$$x=-14$$

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 -14 mō te x_{2}.

$$x^{2}+10x-56=\left(x-4\right)\left(x-\left(-14\right)\right)$$

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

$$x^{2}+10x-56=\left(x-4\right)\left(x+14\right)$$

Aromātai

$$\left(x-4\right)\left(x+14\right)$$