Wolfram|Alpha gives me $-frac14 (1+sqrt5-2 x) (-1+sqrt5+2 x)$.Cyptcouncil.net gives me $(x-frac1+sqrt52)(x-frac1-sqrt52)$.

You are watching: Factor x^2-x+1

The closest ns can acquire is $(x+1)(x-1)-x$.

So exactly how do I gain a pretty answer choose the ones noted above?

Complete the square: conference $x^2-x$ and also whatever continuous you require to develop something that the form $(x-c)^2$, then repair the alters you"ve made:$$ extstyle x^2-x-1 = left( x^2-x+frac14 ight) - frac14-1 = (x-frac12)^2-frac54$$Now the RHS has the type $a^2-b^2$ which friend can element as $(a+b)(a-b)$:$$ extstyle (x-frac12)^2-frac54 = (x-frac12)^2-(fracsqrt52)^2=(x-frac12+fracsqrt52)(x-frac12-fracsqrt52)$$

Solution 1: If $p$ is a source of $f(x)=x^2-x-1$ climate $x-p$ is a factor of $f$ (see https://en.wikipedia.org/wiki/Polynomial_remainder_theorem). Therefore $f(x)=(x-a)(x-b)$ where $a,b$ are the roots of $f$ (given through the quadratic formula). This gives Cyptcouncil.net"s answer.

If you clean the denominators in Cyptcouncil.net"s answer, you obtain Wolfram"s answer.

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

which is a distinction of squares, for this reason it factors as $(x-1/2-sqrt 5/2)(x-1/2+sqrt5/2)$. This is Cyptcouncil.net"s answer.

Apply quadratic formula for the root of $x^2-x-1=0$ as complies with $$x=frac-(-1)pmsqrt(-1)^2-4(1)(-1)2(1)=frac1pm sqrt 52$$hence, one should have actually the following determinants $$x^2-x-1=1cdot left(x-frac1+ sqrt 52 ight)left(x-frac1- sqrt 52 ight)$$or $$frac14(2x-1-sqrt 5)(2x-1+sqrt 5)$$$$=-frac14(1+sqrt 5-2x)(-1+sqrt 5+2x)$$So both the answers are correct

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