Solve (x-2-sqrt{3})(x-2+sqrt{3})=?

$$( x - 2 - \sqrt { 3 } ) ( x - 2 + \sqrt { 3 } ) = ?$$

Evaluate

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

Solution Steps

Apply the distributive property by multiplying each term of x-2-\sqrt{3} by each term of x-2+\sqrt{3}.

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

Combine -2x and -2x to get -4x.

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

Combine x\sqrt{3} and -\sqrt{3}x to get 0.

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

Combine -2\sqrt{3} and 2\sqrt{3} to get 0.

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

The square of \sqrt{3} is 3.

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

Subtract 3 from 4 to get 1.

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

Differentiate w.r.t. x

$$2\left(x-2\right)$$

Steps Using Derivative Rule for Sum

Apply the distributive property by multiplying each term of x-2-\sqrt{3} by each term of x-2+\sqrt{3}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-2x+x\sqrt{3}-2x+4-2\sqrt{3}-\sqrt{3}x+2\sqrt{3}-\left(\sqrt{3}\right)^{2})$$

Combine -2x and -2x to get -4x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-4x+x\sqrt{3}+4-2\sqrt{3}-\sqrt{3}x+2\sqrt{3}-\left(\sqrt{3}\right)^{2})$$

Combine x\sqrt{3} and -\sqrt{3}x to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-4x+4-2\sqrt{3}+2\sqrt{3}-\left(\sqrt{3}\right)^{2})$$

Combine -2\sqrt{3} and 2\sqrt{3} to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-4x+4-\left(\sqrt{3}\right)^{2})$$

The square of \sqrt{3} is 3.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-4x+4-3)$$

Subtract 3 from 4 to get 1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-4x+1)$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

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

Subtract 1 from 2.

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

Subtract 1 from 1.

$$2x^{1}-4x^{0}$$

For any term t, t^{1}=t.

$$2x-4x^{0}$$

For any term t except 0, t^{0}=1.

$$2x-4$$