Solve x^2+4x+3quad=quad(x+1)(x+3)

$$x ^ { 2 } + 4 x + 3 \quad = \quad ( x + 1 ) ( x + 3 )$$

Solve for x (complex solution)

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

Solution Steps

Use the distributive property to multiply x+1 by x+3 and combine like terms.

$$x^{2}+4x+3=x^{2}+4x+3$$

Subtract x^{2} from both sides.

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

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

$$4x+3=4x+3$$

Subtract 4x from both sides.

$$4x+3-4x=3$$

Combine 4x and -4x to get 0.

$$3=3$$

Compare 3 and 3.

$$\text{true}$$

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+1 by x+3 and combine like terms.

$$x^{2}+4x+3=x^{2}+4x+3$$

Subtract x^{2} from both sides.

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

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

$$4x+3=4x+3$$

Subtract 4x from both sides.

$$4x+3-4x=3$$

Combine 4x and -4x to get 0.

$$3=3$$

Compare 3 and 3.

$$\text{true}$$

This is true for any x.

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