Solve (4a+3)(a+5)+(a+6)^2+(a^2-36)

$$( 4 a + 3 ) ( a + 5 ) + ( a + 6 ) ^ { 2 } + ( a ^ { 2 } - 36 )$$

Evaluate

$$6a^{2}+35a+15$$

Solution Steps

Use the distributive property to multiply 4a+3 by a+5 and combine like terms.

$$4a^{2}+23a+15+\left(a+6\right)^{2}+a^{2}-36$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+6\right)^{2}.

$$4a^{2}+23a+15+a^{2}+12a+36+a^{2}-36$$

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

$$5a^{2}+23a+15+12a+36+a^{2}-36$$

Combine 23a and 12a to get 35a.

$$5a^{2}+35a+15+36+a^{2}-36$$

Add 15 and 36 to get 51.

$$5a^{2}+35a+51+a^{2}-36$$

Combine 5a^{2} and a^{2} to get 6a^{2}.

$$6a^{2}+35a+51-36$$

Subtract 36 from 51 to get 15.

$$6a^{2}+35a+15$$

Expand

$$6a^{2}+35a+15$$

Solution Steps

Use the distributive property to multiply 4a+3 by a+5 and combine like terms.

$$4a^{2}+23a+15+\left(a+6\right)^{2}+a^{2}-36$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+6\right)^{2}.

$$4a^{2}+23a+15+a^{2}+12a+36+a^{2}-36$$

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

$$5a^{2}+23a+15+12a+36+a^{2}-36$$

Combine 23a and 12a to get 35a.

$$5a^{2}+35a+15+36+a^{2}-36$$

Add 15 and 36 to get 51.

$$5a^{2}+35a+51+a^{2}-36$$

Combine 5a^{2} and a^{2} to get 6a^{2}.

$$6a^{2}+35a+51-36$$

Subtract 36 from 51 to get 15.

$$6a^{2}+35a+15$$