How do you factor $x^2-x-1$?

Question

I know you can't have all integers, but how do you factor this anyway?

Wolfram|Alpha gives me $-\frac{1}{4} (1+\sqrt{5}-2 x) (-1+\sqrt{5}+2 x)$.
Cymath gives me $(x-\frac{1+\sqrt{5}}{2})(x-\frac{1-\sqrt{5}}{2})$.

The closest I can get is $(x+1)(x-1)-x$.

So how do I get a nice answer like the ones listed above?

Answer 1

Complete the square: Gather $x^2-x$ and whatever constant you need to create something of the form $(x-c)^2$, then repair the changes you've made: $$ \textstyle x^2-x-1 = \left( x^2-x+\frac14 \right) - \frac14-1 = (x-\frac12)^2-\frac54 $$ Now the RHS has the form $a^2-b^2$ which you can factor as $(a+b)(a-b)$: $$ \textstyle (x-\frac12)^2-\frac54 = (x-\frac12)^2-(\frac{\sqrt5}2)^2 =(x-\frac12+\frac{\sqrt5}2)(x-\frac12-\frac{\sqrt5}2) $$

Answer 2

Solution 1: If $p$ is a root of $f(x)=x^2-x-1$ then $x-p$ is a factor of $f$ (see https://en.wikipedia.org/wiki/Polynomial_remainder_theorem). So $f(x)=(x-a)(x-b)$ where $a,b$ are the roots of $f$ (given by the quadratic formula). This gives Cymath's answer.

If you clear the denominators in Cymath's answer, you get Wolfram's answer.

Solution 2: Complete the square. $x^2-x-1\\=x^2-x+1/4-1-1/4 \\= (x-1/2)^2-5/4$,

which is a difference of squares, so it factors as $(x-1/2-\sqrt {5}/2)(x-1/2+\sqrt5/2)$. This is Cymath's answer.

Answer 3

Apply quadratic formula for the roots of $x^2-x-1=0$ as follows $$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)}=\frac{1\pm \sqrt 5}{2}$$ hence, one should have the following factors $$x^2-x-1=1\cdot \left(x-\frac{1+ \sqrt 5}{2}\right)\left(x-\frac{1- \sqrt 5}{2}\right)$$ or $$\frac{1}{4}(2x-1-\sqrt 5)(2x-1+\sqrt 5)$$ $$=-\frac{1}{4}(1+\sqrt 5-2x)(-1+\sqrt 5+2x)$$ So both the answers are correct