Whakaoti i te Q(x)=2x^2-5x-3

$$Q ( x ) = 2 x ^ { 2 } - 5 x - 3$$

Tauwehe

$$\left(x-3\right)\left(2x+1\right)$$

Ngā Upane Otinga

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

$$a+b=-5 ab=2\left(-3\right)=-6$$

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 -6.

$$1,-6 2,-3$$

Tātaihia te tapeke mō ia takirua.

$$1-6=-5 2-3=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -5.

$$a=-6 b=1$$

Tuhia anō te 2x^{2}-5x-3 hei \left(2x^{2}-6x\right)+\left(x-3\right).

$$\left(2x^{2}-6x\right)+\left(x-3\right)$$

Whakatauwehea atu 2x i te 2x^{2}-6x.

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

Whakatauwehea atu te kīanga pātahi x-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-3\right)\left(2x+1\right)$$

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.

$$2x^{2}-5x-3=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\left(-3\right)}}{2\times 2}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-3\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\left(-3\right)}}{2\times 2}$$

Whakareatia -8 ki te -3.

$$x=\frac{-\left(-5\right)±\sqrt{25+24}}{2\times 2}$$

Tāpiri 25 ki te 24.

$$x=\frac{-\left(-5\right)±\sqrt{49}}{2\times 2}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-\left(-5\right)±7}{2\times 2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±7}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{5±7}{4}$$

Nā, me whakaoti te whārite x=\frac{5±7}{4} ina he tāpiri te ±. Tāpiri 5 ki te 7.

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

Whakawehe 12 ki te 4.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{5±7}{4} ina he tango te ±. Tango 7 mai i 5.

$$x=-\frac{2}{4}$$

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

$$x=-\frac{1}{2}$$

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

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

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

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

Tāpiri \frac{1}{2} ki te x mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

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

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

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

Aromātai

$$\left(x-3\right)\left(2x+1\right)$$