Solve (2x-y)(y+2x)-2(3x-2y)(-2y-3x)-(11x-3y)(2x-3y)

$$( 2 x - y ) ( y + 2 x ) - 2 ( 3 x - 2 y ) ( - 2 y - 3 x ) - ( 11 x - 3 y ) ( 2 x - 3 y )$$

Evaluate

$$3y\left(13x-6y\right)$$

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}.

$$\left(2x\right)^{2}-y^{2}-2\left(3x-2y\right)\left(-2y-3x\right)-\left(11x-3y\right)\left(2x-3y\right)$$

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

$$2^{2}x^{2}-y^{2}-2\left(3x-2y\right)\left(-2y-3x\right)-\left(11x-3y\right)\left(2x-3y\right)$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-y^{2}-2\left(3x-2y\right)\left(-2y-3x\right)-\left(11x-3y\right)\left(2x-3y\right)$$

Apply the distributive property by multiplying each term of 11x-3y by each term of 2x-3y.

$$4x^{2}-y^{2}-2\left(3x-2y\right)\left(-2y-3x\right)-\left(22x^{2}-33xy-6yx+9y^{2}\right)$$

Combine -33xy and -6yx to get -39xy.

$$4x^{2}-y^{2}-2\left(3x-2y\right)\left(-2y-3x\right)-\left(22x^{2}-39xy+9y^{2}\right)$$

To find the opposite of 22x^{2}-39xy+9y^{2}, find the opposite of each term.

$$4x^{2}-y^{2}-2\left(3x-2y\right)\left(-2y-3x\right)-22x^{2}-\left(-39xy\right)-9y^{2}$$

The opposite of -39xy is 39xy.

$$4x^{2}-y^{2}-2\left(3x-2y\right)\left(-2y-3x\right)-22x^{2}+39xy-9y^{2}$$

Use the distributive property to multiply -2 by 3x-2y.

$$4x^{2}-y^{2}+\left(-6x+4y\right)\left(-2y-3x\right)-22x^{2}+39xy-9y^{2}$$

Apply the distributive property by multiplying each term of -6x+4y by each term of -2y-3x.

$$4x^{2}-y^{2}+12xy+18x^{2}-8y^{2}-12yx-22x^{2}+39xy-9y^{2}$$

Combine 12xy and -12yx to get 0.

$$4x^{2}-y^{2}+18x^{2}-8y^{2}-22x^{2}+39xy-9y^{2}$$

Combine 4x^{2} and 18x^{2} to get 22x^{2}.

$$22x^{2}-y^{2}-8y^{2}-22x^{2}+39xy-9y^{2}$$

Combine -y^{2} and -8y^{2} to get -9y^{2}.

$$22x^{2}-9y^{2}-22x^{2}+39xy-9y^{2}$$

Combine 22x^{2} and -22x^{2} to get 0.

$$-9y^{2}+39xy-9y^{2}$$

Combine -9y^{2} and -9y^{2} to get -18y^{2}.

$$-18y^{2}+39xy$$

Expand

$$39xy-18y^{2}$$

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}.

$$\left(2x\right)^{2}-y^{2}-2\left(3x-2y\right)\left(-2y-3x\right)-\left(11x-3y\right)\left(2x-3y\right)$$

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

$$2^{2}x^{2}-y^{2}-2\left(3x-2y\right)\left(-2y-3x\right)-\left(11x-3y\right)\left(2x-3y\right)$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-y^{2}-2\left(3x-2y\right)\left(-2y-3x\right)-\left(11x-3y\right)\left(2x-3y\right)$$

Apply the distributive property by multiplying each term of 11x-3y by each term of 2x-3y.

$$4x^{2}-y^{2}-2\left(3x-2y\right)\left(-2y-3x\right)-\left(22x^{2}-33xy-6yx+9y^{2}\right)$$

Combine -33xy and -6yx to get -39xy.

$$4x^{2}-y^{2}-2\left(3x-2y\right)\left(-2y-3x\right)-\left(22x^{2}-39xy+9y^{2}\right)$$

To find the opposite of 22x^{2}-39xy+9y^{2}, find the opposite of each term.

$$4x^{2}-y^{2}-2\left(3x-2y\right)\left(-2y-3x\right)-22x^{2}-\left(-39xy\right)-9y^{2}$$

The opposite of -39xy is 39xy.

$$4x^{2}-y^{2}-2\left(3x-2y\right)\left(-2y-3x\right)-22x^{2}+39xy-9y^{2}$$

Use the distributive property to multiply -2 by 3x-2y.

$$4x^{2}-y^{2}+\left(-6x+4y\right)\left(-2y-3x\right)-22x^{2}+39xy-9y^{2}$$

Apply the distributive property by multiplying each term of -6x+4y by each term of -2y-3x.

$$4x^{2}-y^{2}+12xy+18x^{2}-8y^{2}-12yx-22x^{2}+39xy-9y^{2}$$

Combine 12xy and -12yx to get 0.

$$4x^{2}-y^{2}+18x^{2}-8y^{2}-22x^{2}+39xy-9y^{2}$$

Combine 4x^{2} and 18x^{2} to get 22x^{2}.

$$22x^{2}-y^{2}-8y^{2}-22x^{2}+39xy-9y^{2}$$

Combine -y^{2} and -8y^{2} to get -9y^{2}.

$$22x^{2}-9y^{2}-22x^{2}+39xy-9y^{2}$$

Combine 22x^{2} and -22x^{2} to get 0.

$$-9y^{2}+39xy-9y^{2}$$

Combine -9y^{2} and -9y^{2} to get -18y^{2}.

$$-18y^{2}+39xy$$