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!

Please be sure to answer the question. Administer details and share her research!

But avoid

Asking for help, clarification, or responding to various other answers.Making statements based upon opinion; earlier them up with referrals or an individual experience.

Use invernessgangshow.netJax to layout equations. Invernessgangshow.netJax reference.

See more: About Parasites Can Be Protozoa, Fungi, Or Multicellular Organisms.

To discover more, watch our advice on writing great answers.