Solve P(x)=x^2-3x+2x^4+6quadQ(x)=x^2-2

$$P ( x ) = x ^ { 2 } - 3 x + 2 x ^ { 4 } + 6 \quad Q ( x ) = x ^ { 2 } - 2$$

Solve for Q

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

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract x^{2} from both sides.

$$-3x+2x^{4}+6Qx=x^{2}-2-x^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$-3x+2x^{4}+6Qx=-2$$

Add 3x to both sides.

$$2x^{4}+6Qx=-2+3x$$

Subtract 2x^{4} from both sides.

$$6Qx=-2+3x-2x^{4}$$

The equation is in standard form.

$$6xQ=-2x^{4}+3x-2$$

Divide both sides by 6x.

$$\frac{6xQ}{6x}=\frac{-2x^{4}+3x-2}{6x}$$

Dividing by 6x undoes the multiplication by 6x.

$$Q=\frac{-2x^{4}+3x-2}{6x}$$

Divide -2+3x-2x^{4} by 6x.

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