Solve (2a+1)^2-(2a-2)(2a+2)-5

$$( 2 a + 1 ) ^ { 2 } - ( 2 a - 2 ) ( 2 a + 2 ) - 5$$

Evaluate

$$4a$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(2a+1\right)^{2}.

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

Consider \left(2a-2\right)\left(2a+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.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

$$4a^{2}+4a+1-4a^{2}+4-5$$

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

$$4a+1+4-5$$

Add 1 and 4 to get 5.

$$4a+5-5$$

Subtract 5 from 5 to get 0.

$$4a$$

Expand

$$4a$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(2a+1\right)^{2}.

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

Consider \left(2a-2\right)\left(2a+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.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

$$4a^{2}+4a+1-4a^{2}+4-5$$

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

$$4a+1+4-5$$

Add 1 and 4 to get 5.

$$4a+5-5$$

Subtract 5 from 5 to get 0.

$$4a$$