Solve (4x-1)*(4x+1)-(4x+1-4)^2+(4-1)^2+2*4=

$$( 4 x - 1 ) \cdot ( 4 x + 1 ) - ( 4 x + 1 - 4 ) ^ { 2 } + ( 4 - 1 ) ^ { 2 } + 2 \cdot 4 =$$

Evaluate

$$24x+7$$

Solution Steps

Consider \left(4x-1\right)\left(4x+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.

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

Expand \left(4x\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$16x^{2}-1-\left(4x+1-4\right)^{2}+\left(4-1\right)^{2}+2\times 4$$

Subtract 4 from 1 to get -3.

$$16x^{2}-1-\left(4x-3\right)^{2}+\left(4-1\right)^{2}+2\times 4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$16x^{2}-1-\left(16x^{2}-24x+9\right)+\left(4-1\right)^{2}+2\times 4$$

To find the opposite of 16x^{2}-24x+9, find the opposite of each term.

$$16x^{2}-1-16x^{2}+24x-9+\left(4-1\right)^{2}+2\times 4$$

Combine 16x^{2} and -16x^{2} to get 0.

$$-1+24x-9+\left(4-1\right)^{2}+2\times 4$$

Subtract 9 from -1 to get -10.

$$-10+24x+\left(4-1\right)^{2}+2\times 4$$

Subtract 1 from 4 to get 3.

$$-10+24x+3^{2}+2\times 4$$

Calculate 3 to the power of 2 and get 9.

$$-10+24x+9+2\times 4$$

Add -10 and 9 to get -1.

$$-1+24x+2\times 4$$

Multiply 2 and 4 to get 8.

$$-1+24x+8$$

Add -1 and 8 to get 7.

$$7+24x$$

Expand

$$24x+7$$

Solution Steps

Consider \left(4x-1\right)\left(4x+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.

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

Expand \left(4x\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$16x^{2}-1-\left(4x+1-4\right)^{2}+\left(4-1\right)^{2}+2\times 4$$

Subtract 4 from 1 to get -3.

$$16x^{2}-1-\left(4x-3\right)^{2}+\left(4-1\right)^{2}+2\times 4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$16x^{2}-1-\left(16x^{2}-24x+9\right)+\left(4-1\right)^{2}+2\times 4$$

To find the opposite of 16x^{2}-24x+9, find the opposite of each term.

$$16x^{2}-1-16x^{2}+24x-9+\left(4-1\right)^{2}+2\times 4$$

Combine 16x^{2} and -16x^{2} to get 0.

$$-1+24x-9+\left(4-1\right)^{2}+2\times 4$$

Subtract 9 from -1 to get -10.

$$-10+24x+\left(4-1\right)^{2}+2\times 4$$

Subtract 1 from 4 to get 3.

$$-10+24x+3^{2}+2\times 4$$

Calculate 3 to the power of 2 and get 9.

$$-10+24x+9+2\times 4$$

Add -10 and 9 to get -1.

$$-1+24x+2\times 4$$

Multiply 2 and 4 to get 8.

$$-1+24x+8$$

Add -1 and 8 to get 7.

$$7+24x$$