Wolfram|Alpha provides me \$-frac14 (1+sqrt5-2 x) (-1+sqrt5+2 x)\$.Cyhunterriverpei.com provides me \$(x-frac1+sqrt52)(x-frac1-sqrt52)\$.

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

So just how do I get a nice answer prefer the ones noted above? Complete the square: gather \$x^2-x\$ and whatever constant you need to produce something of the type \$(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 root of \$f(x)=x^2-x-1\$ climate \$x-p\$ is a aspect of \$f\$ (see https://en.wikipedia.org/wiki/Polynomial_remainder_theorem). For this reason \$f(x)=(x-a)(x-b)\$ where \$a,b\$ room the root of \$f\$ (given by the quadratic formula). This offers Cyhunterriverpei.com"s answer.

If you clean the denominators in Cyhunterriverpei.com"s answer, you gain 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, therefore it determinants as \$(x-1/2-sqrt 5/2)(x-1/2+sqrt5/2)\$. This is Cyhunterriverpei.com"s answer. Apply quadratic formula because that the roots of \$x^2-x-1=0\$ as adheres to \$\$x=frac-(-1)pmsqrt(-1)^2-4(1)(-1)2(1)=frac1pm sqrt 52\$\$hence, one should have the following components \$\$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 space correct Thanks for contributing an answer to hunterriverpei.com Stack Exchange!

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