Solve (2x+3)(2x-3)-4x(x-1)+left(x-2right)^2

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

Evaluate

$$x^{2}-5$$

Solution Steps

Consider \left(2x+3\right)\left(2x-3\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 3.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply -4x by x-1.

$$4x^{2}-9-4x^{2}+4x+x^{2}-4x+4$$

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

$$-9+4x+x^{2}-4x+4$$

Combine 4x and -4x to get 0.

$$-9+x^{2}+4$$

Add -9 and 4 to get -5.

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

Expand

$$x^{2}-5$$

Solution Steps

Consider \left(2x+3\right)\left(2x-3\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 3.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply -4x by x-1.

$$4x^{2}-9-4x^{2}+4x+x^{2}-4x+4$$

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

$$-9+4x+x^{2}-4x+4$$

Combine 4x and -4x to get 0.

$$-9+x^{2}+4$$

Add -9 and 4 to get -5.

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