Whakaoti i te operatorname{Cf}(-4)=(-4)^2+3(-4)+11/-4+1

$$\operatorname { Cf } ( - 4 ) = \frac { ( - 4 ) ^ { 2 } + 3 ( - 4 ) + 11 } { - 4 + 1 }$$

Whakaoti mō C

$$C=\frac{5}{4f}$$

$$f\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tātaihia te -4 mā te pū o 2, kia riro ko 16.

$$Cf\left(-4\right)=\frac{16+3\left(-4\right)+11}{-4+1}$$

Whakareatia te 3 ki te -4, ka -12.

$$Cf\left(-4\right)=\frac{16-12+11}{-4+1}$$

Tangohia te 12 i te 16, ka 4.

$$Cf\left(-4\right)=\frac{4+11}{-4+1}$$

Tāpirihia te 4 ki te 11, ka 15.

$$Cf\left(-4\right)=\frac{15}{-4+1}$$

Tāpirihia te -4 ki te 1, ka -3.

$$Cf\left(-4\right)=\frac{15}{-3}$$

Whakawehea te 15 ki te -3, kia riro ko -5.

$$Cf\left(-4\right)=-5$$

He hanga arowhānui tō te whārite.

$$\left(-4f\right)C=-5$$

Whakawehea ngā taha e rua ki te -4f.

$$\frac{\left(-4f\right)C}{-4f}=-\frac{5}{-4f}$$

Mā te whakawehe ki te -4f ka wetekia te whakareanga ki te -4f.

$$C=-\frac{5}{-4f}$$

Whakawehe -5 ki te -4f.

$$C=\frac{5}{4f}$$

Whakaoti mō f

$$f=\frac{5}{4C}$$

$$C\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tātaihia te -4 mā te pū o 2, kia riro ko 16.

$$Cf\left(-4\right)=\frac{16+3\left(-4\right)+11}{-4+1}$$

Whakareatia te 3 ki te -4, ka -12.

$$Cf\left(-4\right)=\frac{16-12+11}{-4+1}$$

Tangohia te 12 i te 16, ka 4.

$$Cf\left(-4\right)=\frac{4+11}{-4+1}$$

Tāpirihia te 4 ki te 11, ka 15.

$$Cf\left(-4\right)=\frac{15}{-4+1}$$

Tāpirihia te -4 ki te 1, ka -3.

$$Cf\left(-4\right)=\frac{15}{-3}$$

Whakawehea te 15 ki te -3, kia riro ko -5.

$$Cf\left(-4\right)=-5$$

He hanga arowhānui tō te whārite.

$$\left(-4C\right)f=-5$$

Whakawehea ngā taha e rua ki te -4C.

$$\frac{\left(-4C\right)f}{-4C}=-\frac{5}{-4C}$$

Mā te whakawehe ki te -4C ka wetekia te whakareanga ki te -4C.

$$f=-\frac{5}{-4C}$$

Whakawehe -5 ki te -4C.

$$f=\frac{5}{4C}$$