Solve 4(x+1)^2-8

$$4 ( x + 1 ) ^ { 2 } - 8$$

Evaluate

$$4\left(\left(x+1\right)^{2}-2\right)$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$4\left(x^{2}+2x+1\right)-8$$

Use the distributive property to multiply 4 by x^{2}+2x+1.

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

Subtract 8 from 4 to get -4.

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

Expand

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

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$4\left(x^{2}+2x+1\right)-8$$

Use the distributive property to multiply 4 by x^{2}+2x+1.

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

Subtract 8 from 4 to get -4.

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