Solve x^4-3x+2=left(x-1right)left(x^3+zx^2+yx-2right)

$${ x }^{ 4 } -3x+2 = \left( x-1 \right) \left( { x }^{ 3 } +z { x }^{ 2 } +yx-2 \right)$$

Solve for y (complex solution)

$$\left\{\begin{matrix}\\y=1+x-xz\text{, }&\text{unconditionally}\\y\in \mathrm{C}\text{, }&x=1\text{ or }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x-1 by x^{3}+zx^{2}+yx-2.

$$x^{4}-3x+2=x^{4}+zx^{3}+yx^{2}-2x-x^{3}-zx^{2}-yx+2$$

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

$$x^{4}+zx^{3}+yx^{2}-2x-x^{3}-zx^{2}-yx+2=x^{4}-3x+2$$

Subtract x^{4} from both sides.

$$zx^{3}+yx^{2}-2x-x^{3}-zx^{2}-yx+2=x^{4}-3x+2-x^{4}$$

Combine x^{4} and -x^{4} to get 0.

$$zx^{3}+yx^{2}-2x-x^{3}-zx^{2}-yx+2=-3x+2$$

Subtract zx^{3} from both sides.

$$yx^{2}-2x-x^{3}-zx^{2}-yx+2=-3x+2-zx^{3}$$

Add 2x to both sides.

$$yx^{2}-x^{3}-zx^{2}-yx+2=-3x+2-zx^{3}+2x$$

Add x^{3} to both sides.

$$yx^{2}-zx^{2}-yx+2=-3x+2-zx^{3}+2x+x^{3}$$

Add zx^{2} to both sides.

$$yx^{2}-yx+2=-3x+2-zx^{3}+2x+x^{3}+zx^{2}$$

Subtract 2 from both sides.

$$yx^{2}-yx=-3x+2-zx^{3}+2x+x^{3}+zx^{2}-2$$

Combine -3x and 2x to get -x.

$$yx^{2}-yx=-x+2-zx^{3}+x^{3}+zx^{2}-2$$

Subtract 2 from 2 to get 0.

$$yx^{2}-yx=-x-zx^{3}+x^{3}+zx^{2}$$

Combine all terms containing y.

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

The equation is in standard form.

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

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

$$\frac{\left(x^{2}-x\right)y}{x^{2}-x}=\frac{x\left(1-x\right)\left(xz-x-1\right)}{x^{2}-x}$$

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

$$y=\frac{x\left(1-x\right)\left(xz-x-1\right)}{x^{2}-x}$$

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

$$y=1+x-xz$$

Solve for y

$$\left\{\begin{matrix}y=1+x-xz\text{, }&x\neq 1\\y\in \mathrm{R}\text{, }&x=1\text{ or }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x-1 by x^{3}+zx^{2}+yx-2.

$$x^{4}-3x+2=x^{4}+zx^{3}+yx^{2}-2x-x^{3}-zx^{2}-yx+2$$

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

$$x^{4}+zx^{3}+yx^{2}-2x-x^{3}-zx^{2}-yx+2=x^{4}-3x+2$$

Subtract x^{4} from both sides.

$$zx^{3}+yx^{2}-2x-x^{3}-zx^{2}-yx+2=x^{4}-3x+2-x^{4}$$

Combine x^{4} and -x^{4} to get 0.

$$zx^{3}+yx^{2}-2x-x^{3}-zx^{2}-yx+2=-3x+2$$

Subtract zx^{3} from both sides.

$$yx^{2}-2x-x^{3}-zx^{2}-yx+2=-3x+2-zx^{3}$$

Add 2x to both sides.

$$yx^{2}-x^{3}-zx^{2}-yx+2=-3x+2-zx^{3}+2x$$

Add x^{3} to both sides.

$$yx^{2}-zx^{2}-yx+2=-3x+2-zx^{3}+2x+x^{3}$$

Add zx^{2} to both sides.

$$yx^{2}-yx+2=-3x+2-zx^{3}+2x+x^{3}+zx^{2}$$

Subtract 2 from both sides.

$$yx^{2}-yx=-3x+2-zx^{3}+2x+x^{3}+zx^{2}-2$$

Combine -3x and 2x to get -x.

$$yx^{2}-yx=-x+2-zx^{3}+x^{3}+zx^{2}-2$$

Subtract 2 from 2 to get 0.

$$yx^{2}-yx=-x-zx^{3}+x^{3}+zx^{2}$$

Combine all terms containing y.

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

The equation is in standard form.

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

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

$$\frac{\left(x^{2}-x\right)y}{x^{2}-x}=\frac{x\left(1-x\right)\left(xz-x-1\right)}{x^{2}-x}$$

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

$$y=\frac{x\left(1-x\right)\left(xz-x-1\right)}{x^{2}-x}$$

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

$$y=1+x-xz$$

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=1\text{; }x=0\text{, }&\text{unconditionally}\\x=-\frac{1-y}{1-z}\text{, }&z\neq 1\\x\in \mathrm{C}\text{, }&y=1\text{ and }z=1\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}\\x=1\text{; }x=0\text{, }&\text{unconditionally}\\x=-\frac{1-y}{1-z}\text{, }&z\neq 1\\x\in \mathrm{R}\text{, }&y=1\text{ and }z=1\end{matrix}\right.$$