Solve left(a+3right)^2-left(a+1right)left(a-1right)-2left(2aright)

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

Evaluate

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

Solution Steps

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

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

Consider \left(a+1\right)\left(a-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

To find the opposite of a^{2}-1, find the opposite of each term.

$$a^{2}+6a+9-a^{2}+1-2\times 2a$$

Combine a^{2} and -a^{2} to get 0.

$$6a+9+1-2\times 2a$$

Add 9 and 1 to get 10.

$$6a+10-2\times 2a$$

Multiply 2 and 2 to get 4.

$$6a+10-4a$$

Combine 6a and -4a to get 2a.

$$2a+10$$

Expand

$$2a+10$$

Solution Steps

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

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

Consider \left(a+1\right)\left(a-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

To find the opposite of a^{2}-1, find the opposite of each term.

$$a^{2}+6a+9-a^{2}+1-2\times 2a$$

Combine a^{2} and -a^{2} to get 0.

$$6a+9+1-2\times 2a$$

Add 9 and 1 to get 10.

$$6a+10-2\times 2a$$

Multiply 2 and 2 to get 4.

$$6a+10-4a$$

Combine 6a and -4a to get 2a.

$$2a+10$$