Solve (2b+2)^2-(2b+2)(2b-2)

$$( 2 b + 2 ) ^ { 2 } - ( 2 b + 2 ) ( 2 b - 2 )$$

Evaluate

$$8\left(b+1\right)$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(2b+2\right)^{2}.

$$4b^{2}+8b+4-\left(2b+2\right)\left(2b-2\right)$$

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

$$4b^{2}+8b+4-\left(\left(2b\right)^{2}-4\right)$$

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

$$4b^{2}+8b+4-\left(2^{2}b^{2}-4\right)$$

Calculate 2 to the power of 2 and get 4.

$$4b^{2}+8b+4-\left(4b^{2}-4\right)$$

To find the opposite of 4b^{2}-4, find the opposite of each term.

$$4b^{2}+8b+4-4b^{2}+4$$

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

$$8b+4+4$$

Add 4 and 4 to get 8.

$$8b+8$$

Expand

$$8b+8$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(2b+2\right)^{2}.

$$4b^{2}+8b+4-\left(2b+2\right)\left(2b-2\right)$$

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

$$4b^{2}+8b+4-\left(\left(2b\right)^{2}-4\right)$$

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

$$4b^{2}+8b+4-\left(2^{2}b^{2}-4\right)$$

Calculate 2 to the power of 2 and get 4.

$$4b^{2}+8b+4-\left(4b^{2}-4\right)$$

To find the opposite of 4b^{2}-4, find the opposite of each term.

$$4b^{2}+8b+4-4b^{2}+4$$

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

$$8b+4+4$$

Add 4 and 4 to get 8.

$$8b+8$$