Solve (2x+y-3)(2x+y+3)-4(x^2+5y^2)

$$( 2 x + y - 3 ) ( 2 x + y + 3 ) - 4 ( x ^ { 2 } + 5 y ^ { 2 } )$$

Evaluate

$$4xy-19y^{2}-9$$

Solution Steps

Consider \left(2x+y-3\right)\left(2x+y+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}, where a=2x+y and b=3. Square 3.

$$\left(2x+y\right)^{2}-9-4\left(x^{2}+5y^{2}\right)$$

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

$$4x^{2}+4xy+y^{2}-9-4\left(x^{2}+5y^{2}\right)$$

Use the distributive property to multiply -4 by x^{2}+5y^{2}.

$$4x^{2}+4xy+y^{2}-9-4x^{2}-20y^{2}$$

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

$$4xy+y^{2}-9-20y^{2}$$

Combine y^{2} and -20y^{2} to get -19y^{2}.

$$4xy-19y^{2}-9$$

Expand

$$4xy-19y^{2}-9$$

Solution Steps

Consider \left(2x+y-3\right)\left(2x+y+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}, where a=2x+y and b=3. Square 3.

$$\left(2x+y\right)^{2}-9-4\left(x^{2}+5y^{2}\right)$$

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

$$4x^{2}+4xy+y^{2}-9-4\left(x^{2}+5y^{2}\right)$$

Use the distributive property to multiply -4 by x^{2}+5y^{2}.

$$4x^{2}+4xy+y^{2}-9-4x^{2}-20y^{2}$$

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

$$4xy+y^{2}-9-20y^{2}$$

Combine y^{2} and -20y^{2} to get -19y^{2}.

$$4xy-19y^{2}-9$$