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

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

Evaluate

$$x^{2}+3$$

Solution Steps

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

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

Consider \left(2x+1\right)\left(2x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+4x+4-1-4x\left(x+1\right)$$

Subtract 1 from 4 to get 3.

$$5x^{2}+4x+3-4x\left(x+1\right)$$

Use the distributive property to multiply -4x by x+1.

$$5x^{2}+4x+3-4x^{2}-4x$$

Combine 5x^{2} and -4x^{2} to get x^{2}.

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

Combine 4x and -4x to get 0.

$$x^{2}+3$$

Expand

$$x^{2}+3$$

Solution Steps

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

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

Consider \left(2x+1\right)\left(2x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+4x+4-1-4x\left(x+1\right)$$

Subtract 1 from 4 to get 3.

$$5x^{2}+4x+3-4x\left(x+1\right)$$

Use the distributive property to multiply -4x by x+1.

$$5x^{2}+4x+3-4x^{2}-4x$$

Combine 5x^{2} and -4x^{2} to get x^{2}.

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

Combine 4x and -4x to get 0.

$$x^{2}+3$$