Solve 25a^2-left(a+3right)^2

$$25 { a }^{ 2 } - { \left(a+3 \right) }^{ 2 }$$

Evaluate

$$24a^{2}-6a-9$$

Solution Steps

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

$$25a^{2}-\left(a^{2}+6a+9\right)$$

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

$$25a^{2}-a^{2}-6a-9$$

Combine 25a^{2} and -a^{2} to get 24a^{2}.

$$24a^{2}-6a-9$$

Expand

$$24a^{2}-6a-9$$

Solution Steps

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

$$25a^{2}-\left(a^{2}+6a+9\right)$$

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

$$25a^{2}-a^{2}-6a-9$$

Combine 25a^{2} and -a^{2} to get 24a^{2}.

$$24a^{2}-6a-9$$