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

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

Evaluate

$$5-4x-3x^{2}$$

Solution Steps

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

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

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

$$-3x^{2}-4x+4+1$$

Add 4 and 1 to get 5.

$$-3x^{2}-4x+5$$

Expand

$$5-4x-3x^{2}$$

Solution Steps

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

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

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

$$-3x^{2}-4x+4+1$$

Add 4 and 1 to get 5.

$$-3x^{2}-4x+5$$