Solve 9left(a+2bright)^2-4left(a-2bright)^2

$$9 { \left(a+2b \right) }^{ 2 } -4 { \left(a-2b \right) }^{ 2 }$$

Evaluate

$$\left(a+10b\right)\left(5a+2b\right)$$

Solution Steps

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

$$9\left(a^{2}+4ab+4b^{2}\right)-4\left(a-2b\right)^{2}$$

Use the distributive property to multiply 9 by a^{2}+4ab+4b^{2}.

$$9a^{2}+36ab+36b^{2}-4\left(a-2b\right)^{2}$$

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

$$9a^{2}+36ab+36b^{2}-4\left(a^{2}-4ab+4b^{2}\right)$$

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

$$9a^{2}+36ab+36b^{2}-4a^{2}+16ab-16b^{2}$$

Combine 9a^{2} and -4a^{2} to get 5a^{2}.

$$5a^{2}+36ab+36b^{2}+16ab-16b^{2}$$

Combine 36ab and 16ab to get 52ab.

$$5a^{2}+52ab+36b^{2}-16b^{2}$$

Combine 36b^{2} and -16b^{2} to get 20b^{2}.

$$5a^{2}+52ab+20b^{2}$$

Expand

$$5a^{2}+52ab+20b^{2}$$

Solution Steps

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

$$9\left(a^{2}+4ab+4b^{2}\right)-4\left(a-2b\right)^{2}$$

Use the distributive property to multiply 9 by a^{2}+4ab+4b^{2}.

$$9a^{2}+36ab+36b^{2}-4\left(a-2b\right)^{2}$$

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

$$9a^{2}+36ab+36b^{2}-4\left(a^{2}-4ab+4b^{2}\right)$$

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

$$9a^{2}+36ab+36b^{2}-4a^{2}+16ab-16b^{2}$$

Combine 9a^{2} and -4a^{2} to get 5a^{2}.

$$5a^{2}+36ab+36b^{2}+16ab-16b^{2}$$

Combine 36ab and 16ab to get 52ab.

$$5a^{2}+52ab+36b^{2}-16b^{2}$$

Combine 36b^{2} and -16b^{2} to get 20b^{2}.

$$5a^{2}+52ab+20b^{2}$$