Solve 3a^2-11a(3b+1)+6(3b+1)^2

$$3 a ^ { 2 } - 11 a ( 3 b + 1 ) + 6 ( 3 b + 1 ) ^ { 2 }$$

Evaluate

$$\left(a-9b-3\right)\left(3a-6b-2\right)$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(3b+1\right)^{2}.

$$3a^{2}-11a\left(3b+1\right)+6\left(9b^{2}+6b+1\right)$$

Use the distributive property to multiply 6 by 9b^{2}+6b+1.

$$3a^{2}-11a\left(3b+1\right)+54b^{2}+36b+6$$

Use the distributive property to multiply -11a by 3b+1.

$$3a^{2}-33ab-11a+54b^{2}+36b+6$$

Expand

$$3a^{2}-33ab-11a+54b^{2}+36b+6$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(3b+1\right)^{2}.

$$3a^{2}-11a\left(3b+1\right)+6\left(9b^{2}+6b+1\right)$$

Use the distributive property to multiply 6 by 9b^{2}+6b+1.

$$3a^{2}-11a\left(3b+1\right)+54b^{2}+36b+6$$

Use the distributive property to multiply -11a by 3b+1.

$$3a^{2}-33ab-11a+54b^{2}+36b+6$$