Solve P(x)=x^3-8x^2-3divQ(x)=x-3

$$P ( x ) = x ^ { 3 } - 8 x ^ { 2 } - 3 \div Q ( x ) = x - 3$$

Solve for Q

$$Q=-\frac{3x}{-x^{3}+8x^{2}+x-3}$$

$$x\neq 0\text{ and }x\neq \frac{\sqrt{67}\left(\sqrt{3}\sin(\frac{\arccos(\frac{1015\sqrt{67}}{8978})}{3})-\cos(\frac{\arccos(\frac{1015\sqrt{67}}{8978})}{3})\right)+8}{3}\text{ and }x\neq \frac{-\sqrt{67}\cos(\frac{\arccos(\frac{1015\sqrt{67}}{8978})}{3})-\sqrt{201}\sin(\frac{\arccos(\frac{1015\sqrt{67}}{8978})}{3})+8}{3}\text{ and }x\neq \frac{2\left(\sqrt{67}\cos(\frac{\arccos(\frac{1015\sqrt{67}}{8978})}{3})+4\right)}{3}$$

Steps for Solving Linear Equation

Variable Q cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by Q.

$$Q\left(x^{3}-8x^{2}\right)-3x=Qx+Q\left(-3\right)$$

Use the distributive property to multiply Q by x^{3}-8x^{2}.

$$Qx^{3}-8Qx^{2}-3x=Qx+Q\left(-3\right)$$

Subtract Qx from both sides.

$$Qx^{3}-8Qx^{2}-3x-Qx=Q\left(-3\right)$$

Subtract Q\left(-3\right) from both sides.

$$Qx^{3}-8Qx^{2}-3x-Qx-Q\left(-3\right)=0$$

Multiply -1 and -3 to get 3.

$$Qx^{3}-8Qx^{2}-3x-Qx+3Q=0$$

Add 3x to both sides. Anything plus zero gives itself.

$$Qx^{3}-8Qx^{2}-Qx+3Q=3x$$

Combine all terms containing Q.

$$\left(x^{3}-8x^{2}-x+3\right)Q=3x$$

Divide both sides by x^{3}-8x^{2}-x+3.

$$\frac{\left(x^{3}-8x^{2}-x+3\right)Q}{x^{3}-8x^{2}-x+3}=\frac{3x}{x^{3}-8x^{2}-x+3}$$

Dividing by x^{3}-8x^{2}-x+3 undoes the multiplication by x^{3}-8x^{2}-x+3.

$$Q=\frac{3x}{x^{3}-8x^{2}-x+3}$$

Variable Q cannot be equal to 0.

$$Q=\frac{3x}{x^{3}-8x^{2}-x+3}\text{, }Q\neq 0$$