Solve (2t+1)^2-(2t+1)(2t-1)

$$( 2 t + 1 ) ^ { 2 } - ( 2 t + 1 ) ( 2 t - 1 )$$

Evaluate

$$4t+2$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2t+1\right)^{2}.

$$4t^{2}+4t+1-\left(2t+1\right)\left(2t-1\right)$$

Consider \left(2t+1\right)\left(2t-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.

$$4t^{2}+4t+1-\left(\left(2t\right)^{2}-1\right)$$

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

$$4t^{2}+4t+1-\left(2^{2}t^{2}-1\right)$$

Calculate 2 to the power of 2 and get 4.

$$4t^{2}+4t+1-\left(4t^{2}-1\right)$$

To find the opposite of 4t^{2}-1, find the opposite of each term.

$$4t^{2}+4t+1-4t^{2}+1$$

Combine 4t^{2} and -4t^{2} to get 0.

$$4t+1+1$$

Add 1 and 1 to get 2.

$$4t+2$$

Expand

$$4t+2$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2t+1\right)^{2}.

$$4t^{2}+4t+1-\left(2t+1\right)\left(2t-1\right)$$

Consider \left(2t+1\right)\left(2t-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.

$$4t^{2}+4t+1-\left(\left(2t\right)^{2}-1\right)$$

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

$$4t^{2}+4t+1-\left(2^{2}t^{2}-1\right)$$

Calculate 2 to the power of 2 and get 4.

$$4t^{2}+4t+1-\left(4t^{2}-1\right)$$

To find the opposite of 4t^{2}-1, find the opposite of each term.

$$4t^{2}+4t+1-4t^{2}+1$$

Combine 4t^{2} and -4t^{2} to get 0.

$$4t+1+1$$

Add 1 and 1 to get 2.

$$4t+2$$