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

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

Evaluate

$$6a-3$$

Solution Steps

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

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

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

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

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

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

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

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

Combine 2a and 4a to get 6a.

$$6a+1-4$$

Subtract 4 from 1 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+1\right)^{2}.

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

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

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

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

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

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

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

Combine 2a and 4a to get 6a.

$$6a+1-4$$

Subtract 4 from 1 to get -3.

$$6a-3$$