Wolfram|Alpha offers me $-\frac14 (1+\sqrt5-2 x) (-1+\sqrt5+2 x)$.Cyinvernessgangshow.net offers me $(x-\frac1+\sqrt52)(x-\frac1-\sqrt52)$.

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

The closest i can obtain is $(x+1)(x-1)-x$.

So how do I obtain a quite answer like the ones listed above? Complete the square: conference $x^2-x$ and whatever continuous you need to produce something of the kind $(x-c)^2$, climate repair the transforms 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 friend can aspect as $(a+b)(a-b)$:$$\textstyle (x-\frac12)^2-\frac54 = (x-\frac12)^2-(\frac\sqrt52)^2=(x-\frac12+\frac\sqrt52)(x-\frac12-\frac\sqrt52)$$ Solution 1: If $p$ is a source of $f(x)=x^2-x-1$ climate $x-p$ is a element of $f$ (see https://en.wikipedia.org/wiki/Polynomial_remainder_theorem). Therefore $f(x)=(x-a)(x-b)$ where $a,b$ room the roots of $f$ (given by the quadratic formula). This offers Cyinvernessgangshow.net"s answer.

If you clear the platform in Cyinvernessgangshow.net"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 distinction of squares, so it components as $(x-1/2-\sqrt 5/2)(x-1/2+\sqrt5/2)$. This is Cyinvernessgangshow.net"s answer. Apply quadratic formula because that the root of $x^2-x-1=0$ as complies with $$x=\frac-(-1)\pm\sqrt(-1)^2-4(1)(-1)2(1)=\frac1\pm \sqrt 52$$hence, one should have the following factors $$x^2-x-1=1\cdot \left(x-\frac1+ \sqrt 52\right)\left(x-\frac1- \sqrt 52\right)$$or $$\frac14(2x-1-\sqrt 5)(2x-1+\sqrt 5)$$$$=-\frac14(1+\sqrt 5-2x)(-1+\sqrt 5+2x)$$So both the answers room correct Thanks for contributing an answer to invernessgangshow.netematics ridge Exchange!

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