Solve x^4+x^2+x+1=(x^2+A)(x^2-1)+Bx+C

$$x ^ { 4 } + x ^ { 2 } + x + 1 = ( x ^ { 2 } + A ) ( x ^ { 2 } - 1 ) + B x + C$$

Solve for A (complex solution)

$$\left\{\begin{matrix}A=-\frac{-2x^{2}+Bx-x+C-1}{x^{2}-1}\text{, }&x\neq -1\text{ and }x\neq 1\\A\in \mathrm{C}\text{, }&\left(B=4-C\text{ and }x=1\right)\text{ or }\left(B=C-2\text{ and }x=-1\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$x^{4}+x^{2}+x+1=x^{4}-x^{2}+Ax^{2}-A+Bx+C$$

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

$$x^{4}-x^{2}+Ax^{2}-A+Bx+C=x^{4}+x^{2}+x+1$$

Subtract x^{4} from both sides.

$$-x^{2}+Ax^{2}-A+Bx+C=x^{4}+x^{2}+x+1-x^{4}$$

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

$$-x^{2}+Ax^{2}-A+Bx+C=x^{2}+x+1$$

Add x^{2} to both sides.

$$Ax^{2}-A+Bx+C=x^{2}+x+1+x^{2}$$

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

$$Ax^{2}-A+Bx+C=2x^{2}+x+1$$

Subtract Bx from both sides.

$$Ax^{2}-A+C=2x^{2}+x+1-Bx$$

Subtract C from both sides.

$$Ax^{2}-A=2x^{2}+x+1-Bx-C$$

Combine all terms containing A.

$$\left(x^{2}-1\right)A=2x^{2}+x+1-Bx-C$$

The equation is in standard form.

$$\left(x^{2}-1\right)A=2x^{2}-Bx+x-C+1$$

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

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

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

$$A=\frac{2x^{2}-Bx+x-C+1}{x^{2}-1}$$

Solve for B (complex solution)

$$\left\{\begin{matrix}B=\frac{1-C+A+x+2x^{2}-Ax^{2}}{x}\text{, }&x\neq 0\\B\in \mathrm{C}\text{, }&A=C-1\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$x^{4}+x^{2}+x+1=x^{4}-x^{2}+Ax^{2}-A+Bx+C$$

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

$$x^{4}-x^{2}+Ax^{2}-A+Bx+C=x^{4}+x^{2}+x+1$$

Subtract x^{4} from both sides.

$$-x^{2}+Ax^{2}-A+Bx+C=x^{4}+x^{2}+x+1-x^{4}$$

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

$$-x^{2}+Ax^{2}-A+Bx+C=x^{2}+x+1$$

Add x^{2} to both sides.

$$Ax^{2}-A+Bx+C=x^{2}+x+1+x^{2}$$

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

$$Ax^{2}-A+Bx+C=2x^{2}+x+1$$

Subtract Ax^{2} from both sides.

$$-A+Bx+C=2x^{2}+x+1-Ax^{2}$$

Add A to both sides.

$$Bx+C=2x^{2}+x+1-Ax^{2}+A$$

Subtract C from both sides.

$$Bx=2x^{2}+x+1-Ax^{2}+A-C$$

Reorder the terms.

$$Bx=-Ax^{2}+2x^{2}+x+A-C+1$$

The equation is in standard form.

$$xB=1-C+A+x+2x^{2}-Ax^{2}$$

Divide both sides by x.

$$\frac{xB}{x}=\frac{1-C+A+x+2x^{2}-Ax^{2}}{x}$$

Dividing by x undoes the multiplication by x.

$$B=\frac{1-C+A+x+2x^{2}-Ax^{2}}{x}$$

Solve for A

$$\left\{\begin{matrix}A=-\frac{-2x^{2}+Bx-x+C-1}{x^{2}-1}\text{, }&|x|\neq 1\\A\in \mathrm{R}\text{, }&\left(B=C-2\text{ and }x=-1\right)\text{ or }\left(B=4-C\text{ and }x=1\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$x^{4}+x^{2}+x+1=x^{4}-x^{2}+Ax^{2}-A+Bx+C$$

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

$$x^{4}-x^{2}+Ax^{2}-A+Bx+C=x^{4}+x^{2}+x+1$$

Subtract x^{4} from both sides.

$$-x^{2}+Ax^{2}-A+Bx+C=x^{4}+x^{2}+x+1-x^{4}$$

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

$$-x^{2}+Ax^{2}-A+Bx+C=x^{2}+x+1$$

Add x^{2} to both sides.

$$Ax^{2}-A+Bx+C=x^{2}+x+1+x^{2}$$

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

$$Ax^{2}-A+Bx+C=2x^{2}+x+1$$

Subtract Bx from both sides.

$$Ax^{2}-A+C=2x^{2}+x+1-Bx$$

Subtract C from both sides.

$$Ax^{2}-A=2x^{2}+x+1-Bx-C$$

Combine all terms containing A.

$$\left(x^{2}-1\right)A=2x^{2}+x+1-Bx-C$$

The equation is in standard form.

$$\left(x^{2}-1\right)A=2x^{2}-Bx+x-C+1$$

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

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

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

$$A=\frac{2x^{2}-Bx+x-C+1}{x^{2}-1}$$

Solve for B

$$\left\{\begin{matrix}B=\frac{1-C+A+x+2x^{2}-Ax^{2}}{x}\text{, }&x\neq 0\\B\in \mathrm{R}\text{, }&A=C-1\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$x^{4}+x^{2}+x+1=x^{4}-x^{2}+Ax^{2}-A+Bx+C$$

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

$$x^{4}-x^{2}+Ax^{2}-A+Bx+C=x^{4}+x^{2}+x+1$$

Subtract x^{4} from both sides.

$$-x^{2}+Ax^{2}-A+Bx+C=x^{4}+x^{2}+x+1-x^{4}$$

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

$$-x^{2}+Ax^{2}-A+Bx+C=x^{2}+x+1$$

Add x^{2} to both sides.

$$Ax^{2}-A+Bx+C=x^{2}+x+1+x^{2}$$

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

$$Ax^{2}-A+Bx+C=2x^{2}+x+1$$

Subtract Ax^{2} from both sides.

$$-A+Bx+C=2x^{2}+x+1-Ax^{2}$$

Add A to both sides.

$$Bx+C=2x^{2}+x+1-Ax^{2}+A$$

Subtract C from both sides.

$$Bx=2x^{2}+x+1-Ax^{2}+A-C$$

Reorder the terms.

$$Bx=-Ax^{2}+2x^{2}+x+A-C+1$$

The equation is in standard form.

$$xB=1-C+A+x+2x^{2}-Ax^{2}$$

Divide both sides by x.

$$\frac{xB}{x}=\frac{1-C+A+x+2x^{2}-Ax^{2}}{x}$$

Dividing by x undoes the multiplication by x.

$$B=\frac{1-C+A+x+2x^{2}-Ax^{2}}{x}$$