Whakaoti i te 1+4x+4x^2

$$1 + 4 x + 4 x ^ { 2 }$$

Tauwehe

$$\left(2x+1\right)^{2}$$

Ngā Upane Otinga

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$4x^{2}+4x+1$$

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 4x^{2}+ax+bx+1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=4 ab=4\times 1=4$$

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 4.

$$1,4 2,2$$

Tātaihia te tapeke mō ia takirua.

$$1+4=5 2+2=4$$

Ko te otinga te takirua ka hoatu i te tapeke 4.

$$a=2 b=2$$

Tuhia anō te 4x^{2}+4x+1 hei \left(4x^{2}+2x\right)+\left(2x+1\right).

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

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

$$2x\left(2x+1\right)+2x+1$$

Whakatauwehea atu te kīanga pātahi 2x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x+1\right)\left(2x+1\right)$$

Tuhia anōtia hei pūrua huarua.

$$\left(2x+1\right)^{2}$$

Ngā Upane Whakamahi Tauwehe o te Pūrua Huarua

Ko te tikanga tātai o tēnei huatoru he pūrua huatoru, ka whakareatia pea e tētahi tauwehe pātahi. Ka taea ngā pūrua huatoru te tauwehe mā te kimi i ngā pūtakerua o ngā kīanga tau ārahi, autō hoki.

$$factor(4x^{2}+4x+1)$$

Kimihia te tauwehe pātahi nui rawa o ngā tau whakarea.

$$gcf(4,4,1)=1$$

Kimihia te pūtakerua o te kīanga tau ārahi, 4x^{2}.

$$\sqrt{4x^{2}}=2x$$

Ko te pūrua huatoru te pūrua o te huarua ko te tapeke tērā, te huatango rānei o ngā pūtakerua o ngā kīanga tau ārahi, autō hoki, e whakaritea ai te tohu e te tohu o te kīanga tau waenga o te pūrua huatoru.

$$\left(2x+1\right)^{2}$$

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.

$$4x^{2}+4x+1=0$$

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.

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

Pūrua 4.

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

Whakareatia -4 ki te 4.

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

Tāpiri 16 ki te -16.

$$x=\frac{-4±\sqrt{0}}{2\times 4}$$

Tuhia te pūtakerua o te 0.

$$x=\frac{-4±0}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-4±0}{8}$$

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

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

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

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

Tāpiri \frac{1}{2} 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.

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

Tāpiri \frac{1}{2} 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.

$$4x^{2}+4x+1=4\times \frac{2x+1}{2}\times \frac{2x+1}{2}$$

Whakareatia \frac{2x+1}{2} ki te \frac{2x+1}{2} 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.

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

Whakareatia 2 ki te 2.

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

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

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

Aromātai

$$\left(2x+1\right)^{2}$$