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

$$( 2 a + b + 3 ) ( 2 a + b - 3 )$$

Evaluate

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

Solution Steps

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}, where a=2a+b and b=3.

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

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

$$4a^{2}+4ab+b^{2}-3^{2}$$

Calculate 3 to the power of 2 and get 9.

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

Expand

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

Solution Steps

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}, where a=2a+b and b=3.

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

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

$$4a^{2}+4ab+b^{2}-3^{2}$$

Calculate 3 to the power of 2 and get 9.

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