Solve (a+2)^2-(a+1)*(-a-1)

$$( a + 2 ) ^ { 2 } - ( a + 1 ) \cdot ( - a - 1 )$$

Evaluate

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

Solution Steps

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

$$a^{2}+4a+4-\left(a+1\right)\left(-a-1\right)$$

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

$$a^{2}+4a+4-\left(a\left(-a\right)-a-a-1\right)$$

To find the opposite of a\left(-a\right)-a-a-1, find the opposite of each term.

$$a^{2}+4a+4-a\left(-a\right)+a-\left(-a\right)+1$$

Multiply -1 and -1 to get 1.

$$a^{2}+4a+4+aa+a-\left(-a\right)+1$$

Multiply a and a to get a^{2}.

$$a^{2}+4a+4+a^{2}+a-\left(-a\right)+1$$

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

$$2a^{2}+4a+4+a-\left(-a\right)+1$$

Combine 4a and a to get 5a.

$$2a^{2}+5a+4-\left(-a\right)+1$$

Add 4 and 1 to get 5.

$$2a^{2}+5a+5-\left(-a\right)$$

Multiply -1 and -1 to get 1.

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

Combine 5a and a to get 6a.

$$2a^{2}+6a+5$$

Expand

$$2a^{2}+6a+5$$

Solution Steps

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

$$a^{2}+4a+4-\left(a+1\right)\left(-a-1\right)$$

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

$$a^{2}+4a+4-\left(a\left(-a\right)-a-a-1\right)$$

To find the opposite of a\left(-a\right)-a-a-1, find the opposite of each term.

$$a^{2}+4a+4-a\left(-a\right)+a-\left(-a\right)+1$$

Multiply -1 and -1 to get 1.

$$a^{2}+4a+4+aa+a-\left(-a\right)+1$$

Multiply a and a to get a^{2}.

$$a^{2}+4a+4+a^{2}+a-\left(-a\right)+1$$

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

$$2a^{2}+4a+4+a-\left(-a\right)+1$$

Combine 4a and a to get 5a.

$$2a^{2}+5a+4-\left(-a\right)+1$$

Add 4 and 1 to get 5.

$$2a^{2}+5a+5-\left(-a\right)$$

Multiply -1 and -1 to get 1.

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

Combine 5a and a to get 6a.

$$2a^{2}+6a+5$$