Solve (t+5)(t-5)-(t-5)^2

$$( t + 5 ) ( t - 5 ) - ( t - 5 ) ^ { 2 }$$

Evaluate

$$10\left(t-5\right)$$

Solution Steps

Consider \left(t+5\right)\left(t-5\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 5.

$$t^{2}-25-\left(t-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-5\right)^{2}.

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

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

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

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

$$-25+10t-25$$

Subtract 25 from -25 to get -50.

$$-50+10t$$

Expand

$$10t-50$$

Solution Steps

Consider \left(t+5\right)\left(t-5\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 5.

$$t^{2}-25-\left(t-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-5\right)^{2}.

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

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

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

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

$$-25+10t-25$$

Subtract 25 from -25 to get -50.

$$-50+10t$$