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

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

Evaluate

$$\left(a-b+7\right)\left(a+b+3\right)$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+5\right)^{2}.

$$a^{2}+10a+25-\left(2-b\right)^{2}$$

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

$$a^{2}+10a+25-\left(4-4b+b^{2}\right)$$

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

$$a^{2}+10a+25-4+4b-b^{2}$$

Subtract 4 from 25 to get 21.

$$a^{2}+10a+21+4b-b^{2}$$

Expand

$$a^{2}+10a-b^{2}+4b+21$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+5\right)^{2}.

$$a^{2}+10a+25-\left(2-b\right)^{2}$$

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

$$a^{2}+10a+25-\left(4-4b+b^{2}\right)$$

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

$$a^{2}+10a+25-4+4b-b^{2}$$

Subtract 4 from 25 to get 21.

$$a^{2}+10a+21+4b-b^{2}$$