Solve {l}{(2x+y)^2+(x-y)(x+y)}{-5x(x-y)}

$$\left. \begin{array} { l } { ( 2 x + y ) ^ { 2 } + ( x - y ) ( x + y ) } \\ { - 5 x ( x - y ) } \end{array} \right.$$

Evaluate

$$9xy$$

Solution Steps

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

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

Consider \left(x-y\right)\left(x+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}.

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

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

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

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

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

Use the distributive property to multiply -5x by x-y.

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

Combine 5x^{2} and -5x^{2} to get 0.

$$4xy+5xy$$

Combine 4xy and 5xy to get 9xy.

$$9xy$$

Expand

$$9xy$$

Solution Steps

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

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

Consider \left(x-y\right)\left(x+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}.

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

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

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

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

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

Use the distributive property to multiply -5x by x-y.

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

Combine 5x^{2} and -5x^{2} to get 0.

$$4xy+5xy$$

Combine 4xy and 5xy to get 9xy.

$$9xy$$