Solve (x+A)(x-3)=x^2+4Bx+(A+B)

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

Solve for A (complex solution)

$$\left\{\begin{matrix}A=\frac{4Bx+3x+B}{x-4}\text{, }&x\neq 4\\A\in \mathrm{C}\text{, }&B=-\frac{12}{17}\text{ and }x=4\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x+A by x-3.

$$x^{2}-3x+Ax-3A=x^{2}+4Bx+A+B$$

Subtract A from both sides.

$$x^{2}-3x+Ax-3A-A=x^{2}+4Bx+B$$

Combine -3A and -A to get -4A.

$$x^{2}-3x+Ax-4A=x^{2}+4Bx+B$$

Subtract x^{2} from both sides.

$$-3x+Ax-4A=x^{2}+4Bx+B-x^{2}$$

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

$$-3x+Ax-4A=4Bx+B$$

Add 3x to both sides.

$$Ax-4A=4Bx+B+3x$$

Combine all terms containing A.

$$\left(x-4\right)A=4Bx+B+3x$$

The equation is in standard form.

$$\left(x-4\right)A=4Bx+3x+B$$

Divide both sides by x-4.

$$\frac{\left(x-4\right)A}{x-4}=\frac{4Bx+3x+B}{x-4}$$

Dividing by x-4 undoes the multiplication by x-4.

$$A=\frac{4Bx+3x+B}{x-4}$$

Solve for B (complex solution)

$$\left\{\begin{matrix}B=-\frac{4A+3x-Ax}{4x+1}\text{, }&x\neq -\frac{1}{4}\\B\in \mathrm{C}\text{, }&A=\frac{3}{17}\text{ and }x=-\frac{1}{4}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x+A by x-3.

$$x^{2}-3x+Ax-3A=x^{2}+4Bx+A+B$$

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

$$x^{2}+4Bx+A+B=x^{2}-3x+Ax-3A$$

Subtract x^{2} from both sides.

$$4Bx+A+B=x^{2}-3x+Ax-3A-x^{2}$$

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

$$4Bx+A+B=-3x+Ax-3A$$

Subtract A from both sides.

$$4Bx+B=-3x+Ax-3A-A$$

Combine -3A and -A to get -4A.

$$4Bx+B=-3x+Ax-4A$$

Combine all terms containing B.

$$\left(4x+1\right)B=-3x+Ax-4A$$

The equation is in standard form.

$$\left(4x+1\right)B=Ax-3x-4A$$

Divide both sides by 4x+1.

$$\frac{\left(4x+1\right)B}{4x+1}=\frac{Ax-3x-4A}{4x+1}$$

Dividing by 4x+1 undoes the multiplication by 4x+1.

$$B=\frac{Ax-3x-4A}{4x+1}$$

Solve for A

$$\left\{\begin{matrix}A=\frac{4Bx+3x+B}{x-4}\text{, }&x\neq 4\\A\in \mathrm{R}\text{, }&B=-\frac{12}{17}\text{ and }x=4\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x+A by x-3.

$$x^{2}-3x+Ax-3A=x^{2}+4Bx+A+B$$

Subtract A from both sides.

$$x^{2}-3x+Ax-3A-A=x^{2}+4Bx+B$$

Combine -3A and -A to get -4A.

$$x^{2}-3x+Ax-4A=x^{2}+4Bx+B$$

Subtract x^{2} from both sides.

$$-3x+Ax-4A=x^{2}+4Bx+B-x^{2}$$

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

$$-3x+Ax-4A=4Bx+B$$

Add 3x to both sides.

$$Ax-4A=4Bx+B+3x$$

Combine all terms containing A.

$$\left(x-4\right)A=4Bx+B+3x$$

The equation is in standard form.

$$\left(x-4\right)A=4Bx+3x+B$$

Divide both sides by x-4.

$$\frac{\left(x-4\right)A}{x-4}=\frac{4Bx+3x+B}{x-4}$$

Dividing by x-4 undoes the multiplication by x-4.

$$A=\frac{4Bx+3x+B}{x-4}$$

Solve for B

$$\left\{\begin{matrix}B=-\frac{4A+3x-Ax}{4x+1}\text{, }&x\neq -\frac{1}{4}\\B\in \mathrm{R}\text{, }&A=\frac{3}{17}\text{ and }x=-\frac{1}{4}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x+A by x-3.

$$x^{2}-3x+Ax-3A=x^{2}+4Bx+A+B$$

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

$$x^{2}+4Bx+A+B=x^{2}-3x+Ax-3A$$

Subtract x^{2} from both sides.

$$4Bx+A+B=x^{2}-3x+Ax-3A-x^{2}$$

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

$$4Bx+A+B=-3x+Ax-3A$$

Subtract A from both sides.

$$4Bx+B=-3x+Ax-3A-A$$

Combine -3A and -A to get -4A.

$$4Bx+B=-3x+Ax-4A$$

Combine all terms containing B.

$$\left(4x+1\right)B=-3x+Ax-4A$$

The equation is in standard form.

$$\left(4x+1\right)B=Ax-3x-4A$$

Divide both sides by 4x+1.

$$\frac{\left(4x+1\right)B}{4x+1}=\frac{Ax-3x-4A}{4x+1}$$

Dividing by 4x+1 undoes the multiplication by 4x+1.

$$B=\frac{Ax-3x-4A}{4x+1}$$