Solve A(x)=x^5+x^2+x+2:B(x)=x^3-x+1

$$A ( x ) = x ^ { 5 } + x ^ { 2 } + x + 2 : B ( x ) = x ^ { 3 } - x + 1$$

Solve for B

$$B=-\frac{2x}{x^{5}-x^{3}+x^{2}+2x-1}$$

$$x\neq 0\text{ and }x\left(x^{4}-x^{2}+x+2\right)\neq 1\text{ and }x^{5}-x^{3}+x^{2}+2x-1\neq 0$$

Steps for Solving Linear Equation

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

$$Bx^{5}+Bx^{2}+Bx+2x=Bx^{3}-xB+B$$

Subtract Bx^{3} from both sides.

$$Bx^{5}+Bx^{2}+Bx+2x-Bx^{3}=-xB+B$$

Add xB to both sides.

$$Bx^{5}+Bx^{2}+Bx+2x-Bx^{3}+xB=B$$

Subtract B from both sides.

$$Bx^{5}+Bx^{2}+Bx+2x-Bx^{3}+xB-B=0$$

Combine Bx and xB to get 2Bx.

$$Bx^{5}+Bx^{2}+2Bx+2x-Bx^{3}-B=0$$

Subtract 2x from both sides. Anything subtracted from zero gives its negation.

$$Bx^{5}+Bx^{2}+2Bx-Bx^{3}-B=-2x$$

Combine all terms containing B.

$$\left(x^{5}+x^{2}+2x-x^{3}-1\right)B=-2x$$

The equation is in standard form.

$$\left(x^{5}-x^{3}+x^{2}+2x-1\right)B=-2x$$

Divide both sides by x^{2}+x^{5}-1+2x-x^{3}.

$$\frac{\left(x^{5}-x^{3}+x^{2}+2x-1\right)B}{x^{5}-x^{3}+x^{2}+2x-1}=-\frac{2x}{x^{5}-x^{3}+x^{2}+2x-1}$$

Dividing by x^{2}+x^{5}-1+2x-x^{3} undoes the multiplication by x^{2}+x^{5}-1+2x-x^{3}.

$$B=-\frac{2x}{x^{5}-x^{3}+x^{2}+2x-1}$$

Variable B cannot be equal to 0.

$$B=-\frac{2x}{x^{5}-x^{3}+x^{2}+2x-1}\text{, }B\neq 0$$