Solve A^3-4A^2-3A+11I=0

$$A ^ { 3 } - 4 A ^ { 2 } - 3 A + 11 I = 0$$

Solve for I

$$I=-\frac{A\left(A^{2}-4A-3\right)}{11}$$

Steps for Solving Linear Equation

Subtract A^{3} from both sides. Anything subtracted from zero gives its negation.

$$-4A^{2}-3A+11I=-A^{3}$$

Add 4A^{2} to both sides.

$$-3A+11I=-A^{3}+4A^{2}$$

Add 3A to both sides.

$$11I=-A^{3}+4A^{2}+3A$$

The equation is in standard form.

$$11I=3A+4A^{2}-A^{3}$$

Divide both sides by 11.

$$\frac{11I}{11}=\frac{A\left(3+4A-A^{2}\right)}{11}$$

Dividing by 11 undoes the multiplication by 11.

$$I=\frac{A\left(3+4A-A^{2}\right)}{11}$$