Whakaoti i te 12k^2+15k-27

$$12 k ^ { 2 } + 15 k - 27$$

Tauwehe

$$3\left(k-1\right)\left(4k+9\right)$$

Ngā Upane Otinga

Tauwehea te 3.

$$3\left(4k^{2}+5k-9\right)$$

Whakaarohia te 4k^{2}+5k-9. Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 4k^{2}+ak+bk-9. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=5 ab=4\left(-9\right)=-36$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Tātaihia te tapeke mō ia takirua.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

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

$$a=-4 b=9$$

Tuhia anō te 4k^{2}+5k-9 hei \left(4k^{2}-4k\right)+\left(9k-9\right).

$$\left(4k^{2}-4k\right)+\left(9k-9\right)$$

Tauwehea te 4k i te tuatahi me te 9 i te rōpū tuarua.

$$4k\left(k-1\right)+9\left(k-1\right)$$

Whakatauwehea atu te kīanga pātahi k-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(k-1\right)\left(4k+9\right)$$

Me tuhi anō te kīanga whakatauwehe katoa.

$$3\left(k-1\right)\left(4k+9\right)$$

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.

$$12k^{2}+15k-27=0$$

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.

$$k=\frac{-15±\sqrt{15^{2}-4\times 12\left(-27\right)}}{2\times 12}$$

Pūrua 15.

$$k=\frac{-15±\sqrt{225-4\times 12\left(-27\right)}}{2\times 12}$$

Whakareatia -4 ki te 12.

$$k=\frac{-15±\sqrt{225-48\left(-27\right)}}{2\times 12}$$

Whakareatia -48 ki te -27.

$$k=\frac{-15±\sqrt{225+1296}}{2\times 12}$$

Tāpiri 225 ki te 1296.

$$k=\frac{-15±\sqrt{1521}}{2\times 12}$$

Tuhia te pūtakerua o te 1521.

$$k=\frac{-15±39}{2\times 12}$$

Whakareatia 2 ki te 12.

$$k=\frac{-15±39}{24}$$

Nā, me whakaoti te whārite k=\frac{-15±39}{24} ina he tāpiri te ±. Tāpiri -15 ki te 39.

$$k=\frac{24}{24}$$

Whakawehe 24 ki te 24.

$$k=1$$

Nā, me whakaoti te whārite k=\frac{-15±39}{24} ina he tango te ±. Tango 39 mai i -15.

$$k=-\frac{54}{24}$$

Whakahekea te hautanga \frac{-54}{24} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$k=-\frac{9}{4}$$

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 1 mō te x_{1} me te -\frac{9}{4} mō te x_{2}.

$$12k^{2}+15k-27=12\left(k-1\right)\left(k-\left(-\frac{9}{4}\right)\right)$$

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

$$12k^{2}+15k-27=12\left(k-1\right)\left(k+\frac{9}{4}\right)$$

Tāpiri \frac{9}{4} ki te k 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.

$$12k^{2}+15k-27=12\left(k-1\right)\times \frac{4k+9}{4}$$

Whakakorea atu te tauwehe pūnoa nui rawa 4 i roto i te 12 me te 4.

$$12k^{2}+15k-27=3\left(k-1\right)\left(4k+9\right)$$

Aromātai

$$3\left(k-1\right)\left(4k+9\right)$$