Solve (a+1)^2+3(a-1)-108

$$( a + 1 ) ^ { 2 } + 3 ( a - 1 ) - 108$$

Evaluate

$$a^{2}+5a-110$$

Solution Steps

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

$$a^{2}+2a+1+3\left(a-1\right)-108$$

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

$$a^{2}+2a+1+3a-3-108$$

Combine 2a and 3a to get 5a.

$$a^{2}+5a+1-3-108$$

Subtract 3 from 1 to get -2.

$$a^{2}+5a-2-108$$

Subtract 108 from -2 to get -110.

$$a^{2}+5a-110$$

Expand

$$a^{2}+5a-110$$

Solution Steps

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

$$a^{2}+2a+1+3\left(a-1\right)-108$$

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

$$a^{2}+2a+1+3a-3-108$$

Combine 2a and 3a to get 5a.

$$a^{2}+5a+1-3-108$$

Subtract 3 from 1 to get -2.

$$a^{2}+5a-2-108$$

Subtract 108 from -2 to get -110.

$$a^{2}+5a-110$$