Solve (x-1-sqrt{2})(x-1+sqrt{2})

$$( x - 1 - \sqrt { 2 } ) ( x - 1 + \sqrt { 2 } )$$

Evaluate

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

Solution Steps

Apply the distributive property by multiplying each term of x-1-\sqrt{2} by each term of x-1+\sqrt{2}.

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

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

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Subtract 2 from 1 to get -1.

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

Differentiate w.r.t. x

$$2\left(x-1\right)$$

Steps Using Derivative Rule for Sum

Apply the distributive property by multiplying each term of x-1-\sqrt{2} by each term of x-1+\sqrt{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-x+x\sqrt{2}-x+1-\sqrt{2}-\sqrt{2}x+\sqrt{2}-\left(\sqrt{2}\right)^{2})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-2x+x\sqrt{2}+1-\sqrt{2}-\sqrt{2}x+\sqrt{2}-\left(\sqrt{2}\right)^{2})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-2x+1-\sqrt{2}+\sqrt{2}-\left(\sqrt{2}\right)^{2})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-2x+1-\left(\sqrt{2}\right)^{2})$$

The square of \sqrt{2} is 2.

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

Subtract 2 from 1 to get -1.

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

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}.

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

Subtract 1 from 2.

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

Subtract 1 from 1.

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

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

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

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

$$2x-2$$