Solve (t+5)^2+(5-2t)(5+2t)-10t

$$( t + 5 ) ^ { 2 } + ( 5 - 2 t ) ( 5 + 2 t ) - 10 t$$

Evaluate

$$50-3t^{2}$$

Solution Steps

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

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

Consider \left(5-2t\right)\left(5+2t\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.

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

Expand \left(2t\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$t^{2}+10t+25+25-4t^{2}-10t$$

Add 25 and 25 to get 50.

$$t^{2}+10t+50-4t^{2}-10t$$

Combine t^{2} and -4t^{2} to get -3t^{2}.

$$-3t^{2}+10t+50-10t$$

Combine 10t and -10t to get 0.

$$-3t^{2}+50$$

Expand

$$50-3t^{2}$$

Solution Steps

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

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

Consider \left(5-2t\right)\left(5+2t\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.

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

Expand \left(2t\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$t^{2}+10t+25+25-4t^{2}-10t$$

Add 25 and 25 to get 50.

$$t^{2}+10t+50-4t^{2}-10t$$

Combine t^{2} and -4t^{2} to get -3t^{2}.

$$-3t^{2}+10t+50-10t$$

Combine 10t and -10t to get 0.

$$-3t^{2}+50$$