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

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

Evaluate

$$2y\left(2x+y\right)$$

Solution Steps

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

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

To find the opposite of 4x^{2}-y^{2}, find the opposite of each term.

$$4x^{2}+4xy+y^{2}-4x^{2}+y^{2}$$

Combine 4x^{2} and -4x^{2} to get 0.

$$4xy+y^{2}+y^{2}$$

Combine y^{2} and y^{2} to get 2y^{2}.

$$4xy+2y^{2}$$

Expand

$$4xy+2y^{2}$$

Solution Steps

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

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

To find the opposite of 4x^{2}-y^{2}, find the opposite of each term.

$$4x^{2}+4xy+y^{2}-4x^{2}+y^{2}$$

Combine 4x^{2} and -4x^{2} to get 0.

$$4xy+y^{2}+y^{2}$$

Combine y^{2} and y^{2} to get 2y^{2}.

$$4xy+2y^{2}$$