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

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

Evaluate

$$6a+3$$

Solution Steps

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

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

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

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

To find the opposite of a^{2}-2a+1, find the opposite of each term.

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

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

$$4a+4+2a-1$$

Combine 4a and 2a to get 6a.

$$6a+4-1$$

Subtract 1 from 4 to get 3.

$$6a+3$$

Expand

$$6a+3$$

Solution Steps

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

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

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

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

To find the opposite of a^{2}-2a+1, find the opposite of each term.

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

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

$$4a+4+2a-1$$

Combine 4a and 2a to get 6a.

$$6a+4-1$$

Subtract 1 from 4 to get 3.

$$6a+3$$