Whakaoti i te f(x)=x^3-6x^2+11x-6g(x)=x+2

$$f ( x ) = x ^ { 3 } - 6 x ^ { 2 } + 11 x - 6 g ( x ) = x + 2$$

Whakaoti mō g

$$g=\frac{x^{2}}{6}-x+\frac{5}{3}-\frac{1}{3x}$$

$$x\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

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

$$-6x^{2}+11x-6gx=x+2-x^{3}$$

Me tāpiri te 6x^{2} ki ngā taha e rua.

$$11x-6gx=x+2-x^{3}+6x^{2}$$

Tangohia te 11x mai i ngā taha e rua.

$$-6gx=x+2-x^{3}+6x^{2}-11x$$

Pahekotia te x me -11x, ka -10x.

$$-6gx=-10x+2-x^{3}+6x^{2}$$

He hanga arowhānui tō te whārite.

$$\left(-6x\right)g=2-10x+6x^{2}-x^{3}$$

Whakawehea ngā taha e rua ki te -6x.

$$\frac{\left(-6x\right)g}{-6x}=\frac{2-10x+6x^{2}-x^{3}}{-6x}$$

Mā te whakawehe ki te -6x ka wetekia te whakareanga ki te -6x.

$$g=\frac{2-10x+6x^{2}-x^{3}}{-6x}$$

Whakawehe -10x+2-x^{3}+6x^{2} ki te -6x.

$$g=\frac{x^{2}}{6}-x+\frac{5}{3}-\frac{1}{3x}$$