Solve (a-5)^2-(a+5)^2=?

$$( a - 5 ) ^ { 2 } - ( a + 5 ) ^ { 2 } = ?$$

Evaluate

$$-20a$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-5\right)^{2}.

$$a^{2}-10a+25-\left(a+5\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+5\right)^{2}.

$$a^{2}-10a+25-\left(a^{2}+10a+25\right)$$

To find the opposite of a^{2}+10a+25, find the opposite of each term.

$$a^{2}-10a+25-a^{2}-10a-25$$

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

$$-10a+25-10a-25$$

Combine -10a and -10a to get -20a.

$$-20a+25-25$$

Subtract 25 from 25 to get 0.

$$-20a$$

Expand

$$-20a$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-5\right)^{2}.

$$a^{2}-10a+25-\left(a+5\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+5\right)^{2}.

$$a^{2}-10a+25-\left(a^{2}+10a+25\right)$$

To find the opposite of a^{2}+10a+25, find the opposite of each term.

$$a^{2}-10a+25-a^{2}-10a-25$$

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

$$-10a+25-10a-25$$

Combine -10a and -10a to get -20a.

$$-20a+25-25$$

Subtract 25 from 25 to get 0.

$$-20a$$