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

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

Evaluate

$$2\left(2a-b\right)\left(4a-b\right)$$

Solution Steps

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

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

Expand \left(2a\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply 3 by 4a^{2}-4ab+b^{2}.

$$4a^{2}-b^{2}+12a^{2}-12ab+3b^{2}$$

Combine 4a^{2} and 12a^{2} to get 16a^{2}.

$$16a^{2}-b^{2}-12ab+3b^{2}$$

Combine -b^{2} and 3b^{2} to get 2b^{2}.

$$16a^{2}+2b^{2}-12ab$$

Expand

$$16a^{2}-12ab+2b^{2}$$

Solution Steps

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

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

Expand \left(2a\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply 3 by 4a^{2}-4ab+b^{2}.

$$4a^{2}-b^{2}+12a^{2}-12ab+3b^{2}$$

Combine 4a^{2} and 12a^{2} to get 16a^{2}.

$$16a^{2}-b^{2}-12ab+3b^{2}$$

Combine -b^{2} and 3b^{2} to get 2b^{2}.

$$16a^{2}+2b^{2}-12ab$$