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

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

Evaluate

$$\left(2a+5\right)\left(3a+4\right)$$

Solution Steps

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

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

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

$$6a^{2}+12a+6+11\left(a+1\right)+3$$

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

$$6a^{2}+12a+6+11a+11+3$$

Combine 12a and 11a to get 23a.

$$6a^{2}+23a+6+11+3$$

Add 6 and 11 to get 17.

$$6a^{2}+23a+17+3$$

Add 17 and 3 to get 20.

$$6a^{2}+23a+20$$

Expand

$$6a^{2}+23a+20$$

Solution Steps

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

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

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

$$6a^{2}+12a+6+11\left(a+1\right)+3$$

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

$$6a^{2}+12a+6+11a+11+3$$

Combine 12a and 11a to get 23a.

$$6a^{2}+23a+6+11+3$$

Add 6 and 11 to get 17.

$$6a^{2}+23a+17+3$$

Add 17 and 3 to get 20.

$$6a^{2}+23a+20$$