Whakaoti i te F(x)=2x^2+x-15

$$F ( x ) = 2 x ^ { 2 } + x - 15$$

Tauwehe

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

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 2x^{2}+ax+bx-15. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=2\left(-15\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Tātaihia te tapeke mō ia takirua.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

Ko te otinga te takirua ka hoatu i te tapeke 1.

$$a=-5 b=6$$

Tuhia anō te 2x^{2}+x-15 hei \left(2x^{2}-5x\right)+\left(6x-15\right).

$$\left(2x^{2}-5x\right)+\left(6x-15\right)$$

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

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

Whakatauwehea atu te kīanga pātahi 2x-5 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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.

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

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.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-15\right)}}{2\times 2}$$

Pūrua 1.

$$x=\frac{-1±\sqrt{1-4\times 2\left(-15\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-1±\sqrt{1-8\left(-15\right)}}{2\times 2}$$

Whakareatia -8 ki te -15.

$$x=\frac{-1±\sqrt{1+120}}{2\times 2}$$

Tāpiri 1 ki te 120.

$$x=\frac{-1±\sqrt{121}}{2\times 2}$$

Tuhia te pūtakerua o te 121.

$$x=\frac{-1±11}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-1±11}{4}$$

Nā, me whakaoti te whārite x=\frac{-1±11}{4} ina he tāpiri te ±. Tāpiri -1 ki te 11.

$$x=\frac{10}{4}$$

Whakahekea te hautanga \frac{10}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{5}{2}$$

Nā, me whakaoti te whārite x=\frac{-1±11}{4} ina he tango te ±. Tango 11 mai i -1.

$$x=-\frac{12}{4}$$

Whakawehe -12 ki te 4.

$$x=-3$$

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 \frac{5}{2} mō te x_{1} me te -3 mō te x_{2}.

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

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

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

Tango \frac{5}{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.

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

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

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

Aromātai

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