Whakaoti i te 8x^2+65x+8

$$8 x ^ { 2 } + 65 x + 8$$

Tauwehe

$$\left(x+8\right)\left(8x+1\right)$$

Ngā Upane Otinga

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

$$a+b=65 ab=8\times 8=64$$

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 64.

$$1,64 2,32 4,16 8,8$$

Tātaihia te tapeke mō ia takirua.

$$1+64=65 2+32=34 4+16=20 8+8=16$$

Ko te otinga te takirua ka hoatu i te tapeke 65.

$$a=1 b=64$$

Tuhia anō te 8x^{2}+65x+8 hei \left(8x^{2}+x\right)+\left(64x+8\right).

$$\left(8x^{2}+x\right)+\left(64x+8\right)$$

Tauwehea te x i te tuatahi me te 8 i te rōpū tuarua.

$$x\left(8x+1\right)+8\left(8x+1\right)$$

Whakatauwehea atu te kīanga pātahi 8x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(8x+1\right)\left(x+8\right)$$

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.

$$8x^{2}+65x+8=0$$

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.

$$x=\frac{-65±\sqrt{65^{2}-4\times 8\times 8}}{2\times 8}$$

Pūrua 65.

$$x=\frac{-65±\sqrt{4225-4\times 8\times 8}}{2\times 8}$$

Whakareatia -4 ki te 8.

$$x=\frac{-65±\sqrt{4225-32\times 8}}{2\times 8}$$

Whakareatia -32 ki te 8.

$$x=\frac{-65±\sqrt{4225-256}}{2\times 8}$$

Tāpiri 4225 ki te -256.

$$x=\frac{-65±\sqrt{3969}}{2\times 8}$$

Tuhia te pūtakerua o te 3969.

$$x=\frac{-65±63}{2\times 8}$$

Whakareatia 2 ki te 8.

$$x=\frac{-65±63}{16}$$

Nā, me whakaoti te whārite x=\frac{-65±63}{16} ina he tāpiri te ±. Tāpiri -65 ki te 63.

$$x=-\frac{2}{16}$$

Whakahekea te hautanga \frac{-2}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{1}{8}$$

Nā, me whakaoti te whārite x=\frac{-65±63}{16} ina he tango te ±. Tango 63 mai i -65.

$$x=-\frac{128}{16}$$

Whakawehe -128 ki te 16.

$$x=-8$$

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}{8} mō te x_{1} me te -8 mō te x_{2}.

$$8x^{2}+65x+8=8\left(x-\left(-\frac{1}{8}\right)\right)\left(x-\left(-8\right)\right)$$

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

$$8x^{2}+65x+8=8\left(x+\frac{1}{8}\right)\left(x+8\right)$$

Tāpiri \frac{1}{8} 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.

$$8x^{2}+65x+8=8\times \frac{8x+1}{8}\left(x+8\right)$$

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

$$8x^{2}+65x+8=\left(8x+1\right)\left(x+8\right)$$

Aromātai

$$\left(x+8\right)\left(8x+1\right)$$