Solve x^3-2x^2+x+2=g(x)(x-2)+(-2x+4)

$$x ^ { 3 } - 2 x ^ { 2 } + x + 2 = g ( x ) ( x - 2 ) + ( - 2 x + 4 )$$

Solve for g

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

$$x\neq 2\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply gx by x-2.

$$x^{3}-2x^{2}+x+2=gx^{2}-2gx-2x+4$$

Swap sides so that all variable terms are on the left hand side.

$$gx^{2}-2gx-2x+4=x^{3}-2x^{2}+x+2$$

Add 2x to both sides.

$$gx^{2}-2gx+4=x^{3}-2x^{2}+x+2+2x$$

Combine x and 2x to get 3x.

$$gx^{2}-2gx+4=x^{3}-2x^{2}+3x+2$$

Subtract 4 from both sides.

$$gx^{2}-2gx=x^{3}-2x^{2}+3x+2-4$$

Subtract 4 from 2 to get -2.

$$gx^{2}-2gx=x^{3}-2x^{2}+3x-2$$

Combine all terms containing g.

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

Divide both sides by x^{2}-2x.

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

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

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

Divide \left(-1+x\right)\left(2-x+x^{2}\right) by x^{2}-2x.

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