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

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

Evaluate

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

Solution Steps

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

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

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

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

Calculate -2 to the power of 2 and get 4.

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

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

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

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

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

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

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

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

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

Expand

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

Solution Steps

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

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

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

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

Calculate -2 to the power of 2 and get 4.

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

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

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

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

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

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

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

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

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