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 Thanks for contributing an answer to ptcouncil.net Stack Exchange!

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