Whakaoti i te 8z^2-10z-3=

$$8 z ^ { 2 } - 10 z - 3 =$$

Tauwehe

$$\left(2z-3\right)\left(4z+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 8z^{2}+az+bz-3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-10 ab=8\left(-3\right)=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Tātaihia te tapeke mō ia takirua.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

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

$$a=-12 b=2$$

Tuhia anō te 8z^{2}-10z-3 hei \left(8z^{2}-12z\right)+\left(2z-3\right).

$$\left(8z^{2}-12z\right)+\left(2z-3\right)$$

Whakatauwehea atu 4z i te 8z^{2}-12z.

$$4z\left(2z-3\right)+2z-3$$

Whakatauwehea atu te kīanga pātahi 2z-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2z-3\right)\left(4z+1\right)$$

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.

$$8z^{2}-10z-3=0$$

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.

$$z=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 8\left(-3\right)}}{2\times 8}$$

Pūrua -10.

$$z=\frac{-\left(-10\right)±\sqrt{100-4\times 8\left(-3\right)}}{2\times 8}$$

Whakareatia -4 ki te 8.

$$z=\frac{-\left(-10\right)±\sqrt{100-32\left(-3\right)}}{2\times 8}$$

Whakareatia -32 ki te -3.

$$z=\frac{-\left(-10\right)±\sqrt{100+96}}{2\times 8}$$

Tāpiri 100 ki te 96.

$$z=\frac{-\left(-10\right)±\sqrt{196}}{2\times 8}$$

Tuhia te pūtakerua o te 196.

$$z=\frac{-\left(-10\right)±14}{2\times 8}$$

Ko te tauaro o -10 ko 10.

$$z=\frac{10±14}{2\times 8}$$

Whakareatia 2 ki te 8.

$$z=\frac{10±14}{16}$$

Nā, me whakaoti te whārite z=\frac{10±14}{16} ina he tāpiri te ±. Tāpiri 10 ki te 14.

$$z=\frac{24}{16}$$

Whakahekea te hautanga \frac{24}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

$$z=\frac{3}{2}$$

Nā, me whakaoti te whārite z=\frac{10±14}{16} ina he tango te ±. Tango 14 mai i 10.

$$z=-\frac{4}{16}$$

Whakahekea te hautanga \frac{-4}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$z=-\frac{1}{4}$$

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

$$8z^{2}-10z-3=8\left(z-\frac{3}{2}\right)\left(z-\left(-\frac{1}{4}\right)\right)$$

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

$$8z^{2}-10z-3=8\left(z-\frac{3}{2}\right)\left(z+\frac{1}{4}\right)$$

Tango \frac{3}{2} mai i z 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.

$$8z^{2}-10z-3=8\times \frac{2z-3}{2}\left(z+\frac{1}{4}\right)$$

Tāpiri \frac{1}{4} ki te z 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.

$$8z^{2}-10z-3=8\times \frac{2z-3}{2}\times \frac{4z+1}{4}$$

Whakareatia \frac{2z-3}{2} ki te \frac{4z+1}{4} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$8z^{2}-10z-3=8\times \frac{\left(2z-3\right)\left(4z+1\right)}{2\times 4}$$

Whakareatia 2 ki te 4.

$$8z^{2}-10z-3=8\times \frac{\left(2z-3\right)\left(4z+1\right)}{8}$$

Whakakorea atu te tauwehe pūnoa nui rawa 8 i roto i te 8 me te 8.

$$8z^{2}-10z-3=\left(2z-3\right)\left(4z+1\right)$$

Aromātai

$$\left(2z-3\right)\left(4z+1\right)$$