Solve {l}{(n+7)^2}{-(n-3)^2}

$$\left. \begin{array} { l } { ( n + 7 ) ^ { 2 } } \\ { - ( n - 3 ) ^ { 2 } } \end{array} \right.$$

Evaluate

$$20\left(n+2\right)$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(n+7\right)^{2}.

$$n^{2}+14n+49-\left(n-3\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-3\right)^{2}.

$$n^{2}+14n+49-\left(n^{2}-6n+9\right)$$

To find the opposite of n^{2}-6n+9, find the opposite of each term.

$$n^{2}+14n+49-n^{2}+6n-9$$

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

$$14n+49+6n-9$$

Combine 14n and 6n to get 20n.

$$20n+49-9$$

Subtract 9 from 49 to get 40.

$$20n+40$$

Expand

$$20n+40$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(n+7\right)^{2}.

$$n^{2}+14n+49-\left(n-3\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-3\right)^{2}.

$$n^{2}+14n+49-\left(n^{2}-6n+9\right)$$

To find the opposite of n^{2}-6n+9, find the opposite of each term.

$$n^{2}+14n+49-n^{2}+6n-9$$

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

$$14n+49+6n-9$$

Combine 14n and 6n to get 20n.

$$20n+49-9$$

Subtract 9 from 49 to get 40.

$$20n+40$$