Solve (2a+b)^2+(a+3)^2

$$( 2 a + b ) ^ { 2 } + ( a + 3 ) ^ { 2 }$$

Expand

$$5a^{2}+4ab+6a+b^{2}+9$$

Solution Steps

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

$$4a^{2}+4ab+b^{2}+\left(a+3\right)^{2}$$

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

$$4a^{2}+4ab+b^{2}+a^{2}+6a+9$$

Combine 4a^{2} and a^{2} to get 5a^{2}.

$$5a^{2}+4ab+b^{2}+6a+9$$

Evaluate

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

Solution Steps

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

$$4a^{2}+4ab+b^{2}+\left(a+3\right)^{2}$$

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

$$4a^{2}+4ab+b^{2}+a^{2}+6a+9$$

Combine 4a^{2} and a^{2} to get 5a^{2}.

$$5a^{2}+4ab+b^{2}+6a+9$$