Solve (x+2)^2-(x+4)^2+4x

$$( x + 2 ) ^ { 2 } - ( x + 4 ) ^ { 2 } + 4 x$$

Evaluate

$$-12$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

$$x^{2}+4x+4-\left(x+4\right)^{2}+4x$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$x^{2}+4x+4-\left(x^{2}+8x+16\right)+4x$$

To find the opposite of x^{2}+8x+16, find the opposite of each term.

$$x^{2}+4x+4-x^{2}-8x-16+4x$$

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

$$4x+4-8x-16+4x$$

Combine 4x and -8x to get -4x.

$$-4x+4-16+4x$$

Subtract 16 from 4 to get -12.

$$-4x-12+4x$$

Combine -4x and 4x to get 0.

$$-12$$

Factor

$$-12$$