Solve (2a+1)^2-(3-a)^2=

$$( 2 a + 1 ) ^ { 2 } - ( 3 - a ) ^ { 2 } =$$

Evaluate

$$\left(3a-2\right)\left(a+4\right)$$

Solution Steps

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

$$4a^{2}+4a+1-\left(3-a\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(3-a\right)^{2}.

$$4a^{2}+4a+1-\left(9-6a+a^{2}\right)$$

To find the opposite of 9-6a+a^{2}, find the opposite of each term.

$$4a^{2}+4a+1-9+6a-a^{2}$$

Subtract 9 from 1 to get -8.

$$4a^{2}+4a-8+6a-a^{2}$$

Combine 4a and 6a to get 10a.

$$4a^{2}+10a-8-a^{2}$$

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

$$3a^{2}+10a-8$$

Expand

$$3a^{2}+10a-8$$

Solution Steps

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

$$4a^{2}+4a+1-\left(3-a\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(3-a\right)^{2}.

$$4a^{2}+4a+1-\left(9-6a+a^{2}\right)$$

To find the opposite of 9-6a+a^{2}, find the opposite of each term.

$$4a^{2}+4a+1-9+6a-a^{2}$$

Subtract 9 from 1 to get -8.

$$4a^{2}+4a-8+6a-a^{2}$$

Combine 4a and 6a to get 10a.

$$4a^{2}+10a-8-a^{2}$$

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

$$3a^{2}+10a-8$$