Whakaoti i te 16x^2+19x+3

$$16 x ^ { 2 } + 19 x + 3$$

Tauwehe

$$\left(x+1\right)\left(16x+3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 16x^{2}+ax+bx+3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=19 ab=16\times 3=48$$

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 48.

$$1,48 2,24 3,16 4,12 6,8$$

Tātaihia te tapeke mō ia takirua.

$$1+48=49 2+24=26 3+16=19 4+12=16 6+8=14$$

Ko te otinga te takirua ka hoatu i te tapeke 19.

$$a=3 b=16$$

Tuhia anō te 16x^{2}+19x+3 hei \left(16x^{2}+3x\right)+\left(16x+3\right).

$$\left(16x^{2}+3x\right)+\left(16x+3\right)$$

Whakatauwehea atu x i te 16x^{2}+3x.

$$x\left(16x+3\right)+16x+3$$

Whakatauwehea atu te kīanga pātahi 16x+3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(16x+3\right)\left(x+1\right)$$

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.

$$16x^{2}+19x+3=0$$

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.

$$x=\frac{-19±\sqrt{19^{2}-4\times 16\times 3}}{2\times 16}$$

Pūrua 19.

$$x=\frac{-19±\sqrt{361-4\times 16\times 3}}{2\times 16}$$

Whakareatia -4 ki te 16.

$$x=\frac{-19±\sqrt{361-64\times 3}}{2\times 16}$$

Whakareatia -64 ki te 3.

$$x=\frac{-19±\sqrt{361-192}}{2\times 16}$$

Tāpiri 361 ki te -192.

$$x=\frac{-19±\sqrt{169}}{2\times 16}$$

Tuhia te pūtakerua o te 169.

$$x=\frac{-19±13}{2\times 16}$$

Whakareatia 2 ki te 16.

$$x=\frac{-19±13}{32}$$

Nā, me whakaoti te whārite x=\frac{-19±13}{32} ina he tāpiri te ±. Tāpiri -19 ki te 13.

$$x=-\frac{6}{32}$$

Whakahekea te hautanga \frac{-6}{32} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{3}{16}$$

Nā, me whakaoti te whārite x=\frac{-19±13}{32} ina he tango te ±. Tango 13 mai i -19.

$$x=-\frac{32}{32}$$

Whakawehe -32 ki te 32.

$$x=-1$$

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}{16} mō te x_{1} me te -1 mō te x_{2}.

$$16x^{2}+19x+3=16\left(x-\left(-\frac{3}{16}\right)\right)\left(x-\left(-1\right)\right)$$

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

$$16x^{2}+19x+3=16\left(x+\frac{3}{16}\right)\left(x+1\right)$$

Tāpiri \frac{3}{16} 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.

$$16x^{2}+19x+3=16\times \frac{16x+3}{16}\left(x+1\right)$$

Whakakorea atu te tauwehe pūnoa nui rawa 16 i roto i te 16 me te 16.

$$16x^{2}+19x+3=\left(16x+3\right)\left(x+1\right)$$

Aromātai

$$\left(x+1\right)\left(16x+3\right)$$