Whakaoti i te x=x(x-a)(x-b)

$$x = x ( x - a ) ( x - b )$$

Whakaoti mō a (complex solution)

$$\left\{\begin{matrix}a=-\frac{x^{2}-bx-1}{b-x}\text{, }&x\neq b\\a\in \mathrm{C}\text{, }&x=0\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te x ki te x-a.

$$x=\left(x^{2}-xa\right)\left(x-b\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}-xa ki te x-b.

$$x=x^{3}-x^{2}b-ax^{2}+axb$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$x^{3}-x^{2}b-ax^{2}+axb=x$$

Tangohia te x^{3} mai i ngā taha e rua.

$$-x^{2}b-ax^{2}+axb=x-x^{3}$$

Me tāpiri te x^{2}b ki ngā taha e rua.

$$-ax^{2}+axb=x-x^{3}+x^{2}b$$

Pahekotia ngā kīanga tau katoa e whai ana i te a.

$$\left(-x^{2}+xb\right)a=x-x^{3}+x^{2}b$$

He hanga arowhānui tō te whārite.

$$\left(bx-x^{2}\right)a=x+bx^{2}-x^{3}$$

Whakawehea ngā taha e rua ki te -x^{2}+xb.

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

Mā te whakawehe ki te -x^{2}+xb ka wetekia te whakareanga ki te -x^{2}+xb.

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

Whakawehe x\left(1-x^{2}+xb\right) ki te -x^{2}+xb.

$$a=\frac{1+bx-x^{2}}{b-x}$$

Whakaoti mō b (complex solution)

$$\left\{\begin{matrix}b=-\frac{x^{2}-ax-1}{a-x}\text{, }&x\neq a\\b\in \mathrm{C}\text{, }&x=0\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te x ki te x-a.

$$x=\left(x^{2}-xa\right)\left(x-b\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}-xa ki te x-b.

$$x=x^{3}-x^{2}b-ax^{2}+xba$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$x^{3}-x^{2}b-ax^{2}+xba=x$$

Tangohia te x^{3} mai i ngā taha e rua.

$$-x^{2}b-ax^{2}+xba=x-x^{3}$$

Me tāpiri te ax^{2} ki ngā taha e rua.

$$-x^{2}b+xba=x-x^{3}+ax^{2}$$

Pahekotia ngā kīanga tau katoa e whai ana i te b.

$$\left(-x^{2}+xa\right)b=x-x^{3}+ax^{2}$$

He hanga arowhānui tō te whārite.

$$\left(ax-x^{2}\right)b=x+ax^{2}-x^{3}$$

Whakawehea ngā taha e rua ki te -x^{2}+xa.

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

Mā te whakawehe ki te -x^{2}+xa ka wetekia te whakareanga ki te -x^{2}+xa.

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

Whakawehe x\left(1-x^{2}+ax\right) ki te -x^{2}+xa.

$$b=\frac{1+ax-x^{2}}{a-x}$$

Whakaoti mō a

$$\left\{\begin{matrix}a=-\frac{x^{2}-bx-1}{b-x}\text{, }&x\neq b\\a\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te x ki te x-a.

$$x=\left(x^{2}-xa\right)\left(x-b\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}-xa ki te x-b.

$$x=x^{3}-x^{2}b-ax^{2}+axb$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$x^{3}-x^{2}b-ax^{2}+axb=x$$

Tangohia te x^{3} mai i ngā taha e rua.

$$-x^{2}b-ax^{2}+axb=x-x^{3}$$

Me tāpiri te x^{2}b ki ngā taha e rua.

$$-ax^{2}+axb=x-x^{3}+x^{2}b$$

Pahekotia ngā kīanga tau katoa e whai ana i te a.

$$\left(-x^{2}+xb\right)a=x-x^{3}+x^{2}b$$

He hanga arowhānui tō te whārite.

$$\left(bx-x^{2}\right)a=x+bx^{2}-x^{3}$$

Whakawehea ngā taha e rua ki te -x^{2}+xb.

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

Mā te whakawehe ki te -x^{2}+xb ka wetekia te whakareanga ki te -x^{2}+xb.

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

Whakawehe x\left(1-x^{2}+xb\right) ki te -x^{2}+xb.

$$a=\frac{1+bx-x^{2}}{b-x}$$

Whakaoti mō b

$$\left\{\begin{matrix}b=-\frac{x^{2}-ax-1}{a-x}\text{, }&x\neq a\\b\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te x ki te x-a.

$$x=\left(x^{2}-xa\right)\left(x-b\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}-xa ki te x-b.

$$x=x^{3}-x^{2}b-ax^{2}+xba$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$x^{3}-x^{2}b-ax^{2}+xba=x$$

Tangohia te x^{3} mai i ngā taha e rua.

$$-x^{2}b-ax^{2}+xba=x-x^{3}$$

Me tāpiri te ax^{2} ki ngā taha e rua.

$$-x^{2}b+xba=x-x^{3}+ax^{2}$$

Pahekotia ngā kīanga tau katoa e whai ana i te b.

$$\left(-x^{2}+xa\right)b=x-x^{3}+ax^{2}$$

He hanga arowhānui tō te whārite.

$$\left(ax-x^{2}\right)b=x+ax^{2}-x^{3}$$

Whakawehea ngā taha e rua ki te -x^{2}+xa.

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

Mā te whakawehe ki te -x^{2}+xa ka wetekia te whakareanga ki te -x^{2}+xa.

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

Whakawehe x\left(1-x^{2}+ax\right) ki te -x^{2}+xa.

$$b=\frac{1+ax-x^{2}}{a-x}$$