Solve x^2-4x-5=(x-5)(x+1)

$${ x }^{ 2 } -4x-5 = (x-5)(x+1)$$

Solve for x (complex solution)

$$x\in \mathrm{C}$$

Solution Steps

Use the distributive property to multiply x-5 by x+1 and combine like terms.

$$x^{2}-4x-5=x^{2}-4x-5$$

Subtract x^{2} from both sides.

$$x^{2}-4x-5-x^{2}=-4x-5$$

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

$$-4x-5=-4x-5$$

Add 4x to both sides.

$$-4x-5+4x=-5$$

Combine -4x and 4x to get 0.

$$-5=-5$$

Compare -5 and -5.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

Use the distributive property to multiply x-5 by x+1 and combine like terms.

$$x^{2}-4x-5=x^{2}-4x-5$$

Subtract x^{2} from both sides.

$$x^{2}-4x-5-x^{2}=-4x-5$$

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

$$-4x-5=-4x-5$$

Add 4x to both sides.

$$-4x-5+4x=-5$$

Combine -4x and 4x to get 0.

$$-5=-5$$

Compare -5 and -5.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$