Whakaoti i te a^3-4*(0)^2+2*(0)-1=

$$a ^ { 3 } - 4 \cdot ( 0 ) ^ { 2 } + 2 \cdot ( 0 ) - 1 =$$

Aromātai

$$a^{3}-1$$

Ngā Upane Otinga

Tātaihia te 0 mā te pū o 2, kia riro ko 0.

$$a^{3}-4\times 0+2\times 0-1$$

Whakareatia te 4 ki te 0, ka 0.

$$a^{3}-0+2\times 0-1$$

Whakareatia te 2 ki te 0, ka 0.

$$a^{3}-0+0-1$$

Tangohia te 1 i te 0, ka -1.

$$a^{3}-0-1$$

Whakareatia te -1 ki te 0, ka 0.

$$a^{3}+0-1$$

Ko te tau i tāpiria he kore ka hua koia tonu.

$$a^{3}-1$$

Tauwehe

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

Ngā Upane Otinga

Whakarea ka paheko i ngā kīanga tau ōrite.

$$a^{3}-1$$

Tuhia anō te a^{3}-1 hei a^{3}-1^{3}. Ka taea te rerekētanga o ngā pūtoru te whakatauwehe mā te whakamahi i te ture: p^{3}-q^{3}=\left(p-q\right)\left(p^{2}+pq+q^{2}\right). Kāore te pūrau a^{2}+a+1 i whakatauwehea i te mea kāhore ōna pūtake whakahau.

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