Solve p(x)=3x^3+x-1quadD(x)=2x^2-1

$$p ( x ) = 3 x ^ { 3 } + x - 1 \quad D ( x ) = 2 x ^ { 2 } - 1$$

Solve for D

$$D=3x^{2}-2x+1+\frac{1}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract 3x^{3} from both sides.

$$x-Dx=2x^{2}-1-3x^{3}$$

Subtract x from both sides.

$$-Dx=2x^{2}-1-3x^{3}-x$$

The equation is in standard form.

$$\left(-x\right)D=-3x^{3}+2x^{2}-x-1$$

Divide both sides by -x.

$$\frac{\left(-x\right)D}{-x}=\frac{-3x^{3}+2x^{2}-x-1}{-x}$$

Dividing by -x undoes the multiplication by -x.

$$D=\frac{-3x^{3}+2x^{2}-x-1}{-x}$$

Divide 2x^{2}-1-3x^{3}-x by -x.

$$D=3x^{2}-2x+1+\frac{1}{x}$$