Whakaoti i te 8x^2-22x+15

$$8 x ^ { 2 } - 22 x + 15$$

Tauwehe

$$\left(2x-3\right)\left(4x-5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 8x^{2}+ax+bx+15. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-22 ab=8\times 15=120$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 120.

$$-1,-120 -2,-60 -3,-40 -4,-30 -5,-24 -6,-20 -8,-15 -10,-12$$

Tātaihia te tapeke mō ia takirua.

$$-1-120=-121 -2-60=-62 -3-40=-43 -4-30=-34 -5-24=-29 -6-20=-26 -8-15=-23 -10-12=-22$$

Ko te otinga te takirua ka hoatu i te tapeke -22.

$$a=-12 b=-10$$

Tuhia anō te 8x^{2}-22x+15 hei \left(8x^{2}-12x\right)+\left(-10x+15\right).

$$\left(8x^{2}-12x\right)+\left(-10x+15\right)$$

Tauwehea te 4x i te tuatahi me te -5 i te rōpū tuarua.

$$4x\left(2x-3\right)-5\left(2x-3\right)$$

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

$$\left(2x-3\right)\left(4x-5\right)$$

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.

$$8x^{2}-22x+15=0$$

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.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 8\times 15}}{2\times 8}$$

Pūrua -22.

$$x=\frac{-\left(-22\right)±\sqrt{484-4\times 8\times 15}}{2\times 8}$$

Whakareatia -4 ki te 8.

$$x=\frac{-\left(-22\right)±\sqrt{484-32\times 15}}{2\times 8}$$

Whakareatia -32 ki te 15.

$$x=\frac{-\left(-22\right)±\sqrt{484-480}}{2\times 8}$$

Tāpiri 484 ki te -480.

$$x=\frac{-\left(-22\right)±\sqrt{4}}{2\times 8}$$

Tuhia te pūtakerua o te 4.

$$x=\frac{-\left(-22\right)±2}{2\times 8}$$

Ko te tauaro o -22 ko 22.

$$x=\frac{22±2}{2\times 8}$$

Whakareatia 2 ki te 8.

$$x=\frac{22±2}{16}$$

Nā, me whakaoti te whārite x=\frac{22±2}{16} ina he tāpiri te ±. Tāpiri 22 ki te 2.

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

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

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

Nā, me whakaoti te whārite x=\frac{22±2}{16} ina he tango te ±. Tango 2 mai i 22.

$$x=\frac{20}{16}$$

Whakahekea te hautanga \frac{20}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x=\frac{5}{4}$$

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{5}{4} mō te x_{2}.

$$8x^{2}-22x+15=8\left(x-\frac{3}{2}\right)\left(x-\frac{5}{4}\right)$$

Tango \frac{3}{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.

$$8x^{2}-22x+15=8\times \frac{2x-3}{2}\left(x-\frac{5}{4}\right)$$

Tango \frac{5}{4} 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.

$$8x^{2}-22x+15=8\times \frac{2x-3}{2}\times \frac{4x-5}{4}$$

Whakareatia \frac{2x-3}{2} ki te \frac{4x-5}{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.

$$8x^{2}-22x+15=8\times \frac{\left(2x-3\right)\left(4x-5\right)}{2\times 4}$$

Whakareatia 2 ki te 4.

$$8x^{2}-22x+15=8\times \frac{\left(2x-3\right)\left(4x-5\right)}{8}$$

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

$$8x^{2}-22x+15=\left(2x-3\right)\left(4x-5\right)$$

Aromātai

$$\left(2x-3\right)\left(4x-5\right)$$