Whakaoti i te {l}{5x^2+10x}{+5}

$$\left. \begin{array} { l } { 5 x ^ { 2 } + 10 x } \\ { + 5 } \end{array} \right.$$

Tauwehe

$$5\left(x+1\right)^{2}$$

Ngā Upane Otinga

Tauwehea te 5.

$$5\left(x^{2}+2x+1\right)$$

Whakaarohia te x^{2}+2x+1. Whakamahia te tikanga tātai pūrua pā, a^{2}+2ab+b^{2}=\left(a+b\right)^{2}, ina a=x, ina b=1.

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

Me tuhi anō te kīanga whakatauwehe katoa.

$$5\left(x+1\right)^{2}$$

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.

$$factor(5x^{2}+10x+5)$$

Kimihia te tauwehe pātahi nui rawa o ngā tau whakarea.

$$gcf(5,10,5)=5$$

Tauwehea te 5.

$$5\left(x^{2}+2x+1\right)$$

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.

$$5\left(x+1\right)^{2}$$

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.

$$5x^{2}+10x+5=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 5\times 5}}{2\times 5}$$

Pūrua 10.

$$x=\frac{-10±\sqrt{100-4\times 5\times 5}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-10±\sqrt{100-20\times 5}}{2\times 5}$$

Whakareatia -20 ki te 5.

$$x=\frac{-10±\sqrt{100-100}}{2\times 5}$$

Tāpiri 100 ki te -100.

$$x=\frac{-10±\sqrt{0}}{2\times 5}$$

Tuhia te pūtakerua o te 0.

$$x=\frac{-10±0}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-10±0}{10}$$

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 -1 mō te x_{2}.

$$5x^{2}+10x+5=5\left(x-\left(-1\right)\right)\left(x-\left(-1\right)\right)$$

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

$$5x^{2}+10x+5=5\left(x+1\right)\left(x+1\right)$$

Aromātai

$$5\left(x+1\right)^{2}$$