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

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

Evaluate

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

Solution Steps

Apply the distributive property by multiplying each term of 3+2a+b by each term of 3-2a+b.

$$9-6a+3b+6a-4a^{2}+2ab+3b-2ba+b^{2}$$

Combine -6a and 6a to get 0.

$$9+3b-4a^{2}+2ab+3b-2ba+b^{2}$$

Combine 3b and 3b to get 6b.

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

Combine 2ab and -2ba to get 0.

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

Expand

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

Solution Steps

Apply the distributive property by multiplying each term of 3+2a+b by each term of 3-2a+b.

$$9-6a+3b+6a-4a^{2}+2ab+3b-2ba+b^{2}$$

Combine -6a and 6a to get 0.

$$9+3b-4a^{2}+2ab+3b-2ba+b^{2}$$

Combine 3b and 3b to get 6b.

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

Combine 2ab and -2ba to get 0.

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