Whakaoti i te F(x)=x^6+9x^3+0+8

$$F ( x ) = x ^ { 6 } + 9 x ^ { 3 } + 0 + 8$$

Tauwehe

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

Ngā Upane Otinga

Kimihia he tauwehe o te āhua x^{k}+m, e wehea ai e x^{k} te huatahi me te pū nui rawa x^{6}, e wehea hoki e m te tauwehe pūmau 8. Ko tētahi tauwehe pērā ko x^{3}+8. Whakatauwehea te pūrau mā te whakawehe ki tēnei tauwehe.

$$\left(x^{3}+8\right)\left(x^{3}+1\right)$$

Whakaarohia te x^{3}+8. Tuhia anō te x^{3}+8 hei x^{3}+2^{3}. Ka taea te tapeke pūtoru te whakatauwehe mā te whakamahi i te ture: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

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

Whakaarohia te x^{3}+1. Tuhia anō te x^{3}+1 hei x^{3}+1^{3}. Ka taea te tapeke pūtoru te whakatauwehe mā te whakamahi i te ture: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

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

Me tuhi anō te kīanga whakatauwehe katoa. Kāore i tauwehea ēnei pūrau i te mea kāhore ō rātou pūtake whakahau: x^{2}-x+1,x^{2}-2x+4.

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

Aromātai

$$x^{6}+9x^{3}+8$$

Ngā Upane Otinga

Tāpirihia te 0 ki te 8, ka 8.

$$x^{6}+9x^{3}+8$$