Solve 4a(4a+3)-(2a+1)(2a-1)

$$4 a ( 4 a + 3 ) - ( 2 a + 1 ) ( 2 a - 1 )$$

Evaluate

$$12a^{2}+12a+1$$

Solution Steps

Use the distributive property to multiply 4a by 4a+3.

$$16a^{2}+12a-\left(2a+1\right)\left(2a-1\right)$$

Consider \left(2a+1\right)\left(2a-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}.

$$16a^{2}+12a-\left(\left(2a\right)^{2}-1^{2}\right)$$

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

$$16a^{2}+12a-\left(2^{2}a^{2}-1^{2}\right)$$

Calculate 2 to the power of 2 and get 4.

$$16a^{2}+12a-\left(4a^{2}-1^{2}\right)$$

Calculate 1 to the power of 2 and get 1.

$$16a^{2}+12a-\left(4a^{2}-1\right)$$

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

$$16a^{2}+12a-4a^{2}-\left(-1\right)$$

The opposite of -1 is 1.

$$16a^{2}+12a-4a^{2}+1$$

Combine 16a^{2} and -4a^{2} to get 12a^{2}.

$$12a^{2}+12a+1$$

Expand

$$12a^{2}+12a+1$$

Solution Steps

Use the distributive property to multiply 4a by 4a+3.

$$16a^{2}+12a-\left(2a+1\right)\left(2a-1\right)$$

Consider \left(2a+1\right)\left(2a-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}.

$$16a^{2}+12a-\left(\left(2a\right)^{2}-1^{2}\right)$$

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

$$16a^{2}+12a-\left(2^{2}a^{2}-1^{2}\right)$$

Calculate 2 to the power of 2 and get 4.

$$16a^{2}+12a-\left(4a^{2}-1^{2}\right)$$

Calculate 1 to the power of 2 and get 1.

$$16a^{2}+12a-\left(4a^{2}-1\right)$$

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

$$16a^{2}+12a-4a^{2}-\left(-1\right)$$

The opposite of -1 is 1.

$$16a^{2}+12a-4a^{2}+1$$

Combine 16a^{2} and -4a^{2} to get 12a^{2}.

$$12a^{2}+12a+1$$