Solve (2x+1)(2x-1)-2(3x+2)(-3x-2)

$$( 2 x + 1 ) ( 2 x - 1 ) - 2 ( 3 x + 2 ) ( - 3 x - 2 )$$

Evaluate

$$22x^{2}+24x+7$$

Solution Steps

Consider \left(2x+1\right)\left(2x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(2x\right)^{2}-1^{2}-2\left(3x+2\right)\left(-3x-2\right)$$

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

$$2^{2}x^{2}-1^{2}-2\left(3x+2\right)\left(-3x-2\right)$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-1^{2}-2\left(3x+2\right)\left(-3x-2\right)$$

Calculate 1 to the power of 2 and get 1.

$$4x^{2}-1-2\left(3x+2\right)\left(-3x-2\right)$$

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

$$4x^{2}-1+\left(-6x-4\right)\left(-3x-2\right)$$

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

$$4x^{2}-1+18x^{2}+12x+12x+8$$

Combine 12x and 12x to get 24x.

$$4x^{2}-1+18x^{2}+24x+8$$

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

$$22x^{2}-1+24x+8$$

Add -1 and 8 to get 7.

$$22x^{2}+7+24x$$

Expand

$$22x^{2}+24x+7$$

Solution Steps

Consider \left(2x+1\right)\left(2x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(2x\right)^{2}-1^{2}-2\left(3x+2\right)\left(-3x-2\right)$$

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

$$2^{2}x^{2}-1^{2}-2\left(3x+2\right)\left(-3x-2\right)$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-1^{2}-2\left(3x+2\right)\left(-3x-2\right)$$

Calculate 1 to the power of 2 and get 1.

$$4x^{2}-1-2\left(3x+2\right)\left(-3x-2\right)$$

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

$$4x^{2}-1+\left(-6x-4\right)\left(-3x-2\right)$$

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

$$4x^{2}-1+18x^{2}+12x+12x+8$$

Combine 12x and 12x to get 24x.

$$4x^{2}-1+18x^{2}+24x+8$$

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

$$22x^{2}-1+24x+8$$

Add -1 and 8 to get 7.

$$22x^{2}+7+24x$$