Solve (2a+1)^2-(a-9)^2

$$( 2 a + 1 ) ^ { 2 } - ( a - 9 ) ^ { 2 }$$

Evaluate

$$\left(3a-8\right)\left(a+10\right)$$

Solution Steps

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

$$4a^{2}+4a+1-\left(a-9\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-9\right)^{2}.

$$4a^{2}+4a+1-\left(a^{2}-18a+81\right)$$

To find the opposite of a^{2}-18a+81, find the opposite of each term.

$$4a^{2}+4a+1-a^{2}+18a-81$$

Combine 4a^{2} and -a^{2} to get 3a^{2}.

$$3a^{2}+4a+1+18a-81$$

Combine 4a and 18a to get 22a.

$$3a^{2}+22a+1-81$$

Subtract 81 from 1 to get -80.

$$3a^{2}+22a-80$$

Expand

$$3a^{2}+22a-80$$

Solution Steps

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

$$4a^{2}+4a+1-\left(a-9\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-9\right)^{2}.

$$4a^{2}+4a+1-\left(a^{2}-18a+81\right)$$

To find the opposite of a^{2}-18a+81, find the opposite of each term.

$$4a^{2}+4a+1-a^{2}+18a-81$$

Combine 4a^{2} and -a^{2} to get 3a^{2}.

$$3a^{2}+4a+1+18a-81$$

Combine 4a and 18a to get 22a.

$$3a^{2}+22a+1-81$$

Subtract 81 from 1 to get -80.

$$3a^{2}+22a-80$$