Solve (2x+1)(2x-1)+4x[2+3(x+1)]

$$( 2 x + 1 ) ( 2 x - 1 ) + 4 x [ 2 + 3 ( x + 1 ) ]$$

Evaluate

$$16x^{2}+20x-1$$

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}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

Calculate 1 to the power of 2 and get 1.

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

Use the distributive property to multiply 3 by x+1.

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

Add 2 and 3 to get 5.

$$4x^{2}-1+4x\left(5+3x\right)$$

Use the distributive property to multiply 4x by 5+3x.

$$4x^{2}-1+20x+12x^{2}$$

Combine 4x^{2} and 12x^{2} to get 16x^{2}.

$$16x^{2}-1+20x$$

Expand

$$16x^{2}+20x-1$$

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}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

Calculate 1 to the power of 2 and get 1.

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

Use the distributive property to multiply 3 by x+1.

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

Add 2 and 3 to get 5.

$$4x^{2}-1+4x\left(5+3x\right)$$

Use the distributive property to multiply 4x by 5+3x.

$$4x^{2}-1+20x+12x^{2}$$

Combine 4x^{2} and 12x^{2} to get 16x^{2}.

$$16x^{2}-1+20x$$