Wolfram|Alpha provides me $-frac14 (1+sqrt5-2 x) (-1+sqrt5+2 x)$.Cyhunterriverpei.com provides me $(x-frac1+sqrt52)(x-frac1-sqrt52)$.
You are watching: Factor x^2+x+1
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
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