Solve =(x-(1-sqrt{5}))(x-(1+sqrt{5}))(x+3)

$$= ( x - ( 1 - \sqrt { 5 } ) ) ( x - ( 1 + \sqrt { 5 } ) ) ( x + 3 )$$

Evaluate

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

Solution Steps

To find the opposite of 1-\sqrt{5}, find the opposite of each term.

$$\left(x-1-\left(-\sqrt{5}\right)\right)\left(x-\left(1+\sqrt{5}\right)\right)\left(x+3\right)$$

The opposite of -\sqrt{5} is \sqrt{5}.

$$\left(x-1+\sqrt{5}\right)\left(x-\left(1+\sqrt{5}\right)\right)\left(x+3\right)$$

To find the opposite of 1+\sqrt{5}, find the opposite of each term.

$$\left(x-1+\sqrt{5}\right)\left(x-1-\sqrt{5}\right)\left(x+3\right)$$

Apply the distributive property by multiplying each term of x-1+\sqrt{5} by each term of x-1-\sqrt{5}.

$$\left(x^{2}-x-x\sqrt{5}-x+1+\sqrt{5}+\sqrt{5}x-\sqrt{5}-\left(\sqrt{5}\right)^{2}\right)\left(x+3\right)$$

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

$$\left(x^{2}-2x-x\sqrt{5}+1+\sqrt{5}+\sqrt{5}x-\sqrt{5}-\left(\sqrt{5}\right)^{2}\right)\left(x+3\right)$$

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

$$\left(x^{2}-2x+1+\sqrt{5}-\sqrt{5}-\left(\sqrt{5}\right)^{2}\right)\left(x+3\right)$$

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

$$\left(x^{2}-2x+1-\left(\sqrt{5}\right)^{2}\right)\left(x+3\right)$$

The square of \sqrt{5} is 5.

$$\left(x^{2}-2x+1-5\right)\left(x+3\right)$$

Subtract 5 from 1 to get -4.

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

Apply the distributive property by multiplying each term of x^{2}-2x-4 by each term of x+3.

$$x^{3}+3x^{2}-2x^{2}-6x-4x-12$$

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

$$x^{3}+x^{2}-6x-4x-12$$

Combine -6x and -4x to get -10x.

$$x^{3}+x^{2}-10x-12$$

Differentiate w.r.t. x

$$3x^{2}+2x-10$$

Steps Using Derivative Rule for Sum

To find the opposite of 1-\sqrt{5}, find the opposite of each term.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(x-1-\left(-\sqrt{5}\right)\right)\left(x-\left(1+\sqrt{5}\right)\right)\left(x+3\right))$$

The opposite of -\sqrt{5} is \sqrt{5}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(x-1+\sqrt{5}\right)\left(x-\left(1+\sqrt{5}\right)\right)\left(x+3\right))$$

To find the opposite of 1+\sqrt{5}, find the opposite of each term.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(x-1+\sqrt{5}\right)\left(x-1-\sqrt{5}\right)\left(x+3\right))$$

Apply the distributive property by multiplying each term of x-1+\sqrt{5} by each term of x-1-\sqrt{5}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(x^{2}-x-x\sqrt{5}-x+1+\sqrt{5}+\sqrt{5}x-\sqrt{5}-\left(\sqrt{5}\right)^{2}\right)\left(x+3\right))$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(x^{2}-2x-x\sqrt{5}+1+\sqrt{5}+\sqrt{5}x-\sqrt{5}-\left(\sqrt{5}\right)^{2}\right)\left(x+3\right))$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(x^{2}-2x+1+\sqrt{5}-\sqrt{5}-\left(\sqrt{5}\right)^{2}\right)\left(x+3\right))$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(x^{2}-2x+1-\left(\sqrt{5}\right)^{2}\right)\left(x+3\right))$$

The square of \sqrt{5} is 5.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(x^{2}-2x+1-5\right)\left(x+3\right))$$

Subtract 5 from 1 to get -4.

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

Apply the distributive property by multiplying each term of x^{2}-2x-4 by each term of x+3.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{3}+3x^{2}-2x^{2}-6x-4x-12)$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{3}+x^{2}-6x-4x-12)$$

Combine -6x and -4x to get -10x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{3}+x^{2}-10x-12)$$

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}.

$$3x^{3-1}+2x^{2-1}-10x^{1-1}$$

Subtract 1 from 3.

$$3x^{2}+2x^{2-1}-10x^{1-1}$$

Subtract 1 from 2.

$$3x^{2}+2x^{1}-10x^{1-1}$$

Subtract 1 from 1.

$$3x^{2}+2x^{1}-10x^{0}$$

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

$$3x^{2}+2x-10x^{0}$$

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

$$3x^{2}+2x-10$$