Whakaoti i te 4x^2+7x+3

$$4 x ^ { 2 } + 7 x + 3$$

Tauwehe

$$\left(x+1\right)\left(4x+3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 4x^{2}+ax+bx+3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=7 ab=4\times 3=12$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 12.

$$1,12 2,6 3,4$$

Tātaihia te tapeke mō ia takirua.

$$1+12=13 2+6=8 3+4=7$$

Ko te otinga te takirua ka hoatu i te tapeke 7.

$$a=3 b=4$$

Tuhia anō te 4x^{2}+7x+3 hei \left(4x^{2}+3x\right)+\left(4x+3\right).

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

Whakatauwehea atu x i te 4x^{2}+3x.

$$x\left(4x+3\right)+4x+3$$

Whakatauwehea atu te kīanga pātahi 4x+3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(4x+3\right)\left(x+1\right)$$

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.

$$4x^{2}+7x+3=0$$

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.

$$x=\frac{-7±\sqrt{7^{2}-4\times 4\times 3}}{2\times 4}$$

Pūrua 7.

$$x=\frac{-7±\sqrt{49-4\times 4\times 3}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-7±\sqrt{49-16\times 3}}{2\times 4}$$

Whakareatia -16 ki te 3.

$$x=\frac{-7±\sqrt{49-48}}{2\times 4}$$

Tāpiri 49 ki te -48.

$$x=\frac{-7±\sqrt{1}}{2\times 4}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-7±1}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-7±1}{8}$$

Nā, me whakaoti te whārite x=\frac{-7±1}{8} ina he tāpiri te ±. Tāpiri -7 ki te 1.

$$x=-\frac{6}{8}$$

Whakahekea te hautanga \frac{-6}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{3}{4}$$

Nā, me whakaoti te whārite x=\frac{-7±1}{8} ina he tango te ±. Tango 1 mai i -7.

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

Whakawehe -8 ki te 8.

$$x=-1$$

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 -\frac{3}{4} mō te x_{1} me te -1 mō te x_{2}.

$$4x^{2}+7x+3=4\left(x-\left(-\frac{3}{4}\right)\right)\left(x-\left(-1\right)\right)$$

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

$$4x^{2}+7x+3=4\left(x+\frac{3}{4}\right)\left(x+1\right)$$

Tāpiri \frac{3}{4} 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.

$$4x^{2}+7x+3=4\times \frac{4x+3}{4}\left(x+1\right)$$

Whakakorea atu te tauwehe pūnoa nui rawa 4 i roto i te 4 me te 4.

$$4x^{2}+7x+3=\left(4x+3\right)\left(x+1\right)$$

Aromātai

$$\left(x+1\right)\left(4x+3\right)$$