Solve (x+b)(x-4)+(-4+x)(x+2)=0

$$( x + b ) ( x - 4 ) + ( - 4 + x ) ( x + 2 ) = 0$$

Solve for b (complex solution)

$$\left\{\begin{matrix}\\b=-2x-2\text{, }&\text{unconditionally}\\b\in \mathrm{C}\text{, }&x=4\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x+b by x-4.

$$x^{2}-4x+bx-4b+\left(-4+x\right)\left(x+2\right)=0$$

Use the distributive property to multiply -4+x by x+2 and combine like terms.

$$x^{2}-4x+bx-4b-2x-8+x^{2}=0$$

Combine -4x and -2x to get -6x.

$$x^{2}-6x+bx-4b-8+x^{2}=0$$

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

$$2x^{2}-6x+bx-4b-8=0$$

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

$$-6x+bx-4b-8=-2x^{2}$$

Add 6x to both sides.

$$bx-4b-8=-2x^{2}+6x$$

Add 8 to both sides.

$$bx-4b=-2x^{2}+6x+8$$

Combine all terms containing b.

$$\left(x-4\right)b=-2x^{2}+6x+8$$

The equation is in standard form.

$$\left(x-4\right)b=8+6x-2x^{2}$$

Divide both sides by x-4.

$$\frac{\left(x-4\right)b}{x-4}=-\frac{2\left(x-4\right)\left(x+1\right)}{x-4}$$

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

$$b=-\frac{2\left(x-4\right)\left(x+1\right)}{x-4}$$

Divide -2\left(-4+x\right)\left(1+x\right) by x-4.

$$b=-2x-2$$

Solve for b

$$\left\{\begin{matrix}\\b=-2x-2\text{, }&\text{unconditionally}\\b\in \mathrm{R}\text{, }&x=4\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x+b by x-4.

$$x^{2}-4x+bx-4b+\left(-4+x\right)\left(x+2\right)=0$$

Use the distributive property to multiply -4+x by x+2 and combine like terms.

$$x^{2}-4x+bx-4b-2x-8+x^{2}=0$$

Combine -4x and -2x to get -6x.

$$x^{2}-6x+bx-4b-8+x^{2}=0$$

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

$$2x^{2}-6x+bx-4b-8=0$$

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

$$-6x+bx-4b-8=-2x^{2}$$

Add 6x to both sides.

$$bx-4b-8=-2x^{2}+6x$$

Add 8 to both sides.

$$bx-4b=-2x^{2}+6x+8$$

Combine all terms containing b.

$$\left(x-4\right)b=-2x^{2}+6x+8$$

The equation is in standard form.

$$\left(x-4\right)b=8+6x-2x^{2}$$

Divide both sides by -4+x.

$$\frac{\left(x-4\right)b}{x-4}=-\frac{2\left(x-4\right)\left(x+1\right)}{x-4}$$

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

$$b=-\frac{2\left(x-4\right)\left(x+1\right)}{x-4}$$

Divide -2\left(-4+x\right)\left(1+x\right) by -4+x.

$$b=-2x-2$$

Solve for x

$$x=-\frac{b}{2}-1$$

$$x=4$$