Solve (x-2-sqrt{5})(x-2+sqrt{5})

$$( x - 2 - \sqrt { 5 } ) ( x - 2 + \sqrt { 5 } )$$

Evaluate

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

Solution Steps

Apply the distributive property by multiplying each term of x-2-\sqrt{5} by each term of x-2+\sqrt{5}.

$$x^{2}-2x+x\sqrt{5}-2x+4-2\sqrt{5}-\sqrt{5}x+2\sqrt{5}-\left(\sqrt{5}\right)^{2}$$

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

$$x^{2}-4x+x\sqrt{5}+4-2\sqrt{5}-\sqrt{5}x+2\sqrt{5}-\left(\sqrt{5}\right)^{2}$$

Combine x\sqrt{5} and -\sqrt{5}x to get 0.

$$x^{2}-4x+4-2\sqrt{5}+2\sqrt{5}-\left(\sqrt{5}\right)^{2}$$

Combine -2\sqrt{5} and 2\sqrt{5} to get 0.

$$x^{2}-4x+4-\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

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

Subtract 5 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{5} by each term of x-2+\sqrt{5}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-2x+x\sqrt{5}-2x+4-2\sqrt{5}-\sqrt{5}x+2\sqrt{5}-\left(\sqrt{5}\right)^{2})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-4x+x\sqrt{5}+4-2\sqrt{5}-\sqrt{5}x+2\sqrt{5}-\left(\sqrt{5}\right)^{2})$$

Combine x\sqrt{5} and -\sqrt{5}x to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-4x+4-2\sqrt{5}+2\sqrt{5}-\left(\sqrt{5}\right)^{2})$$

Combine -2\sqrt{5} and 2\sqrt{5} to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-4x+4-\left(\sqrt{5}\right)^{2})$$

The square of \sqrt{5} is 5.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-4x+4-5)$$

Subtract 5 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.

$$2x-4$$