I want to derive a polynomial equation from the following roots: $$\left\{-1,1+\sqrt5,1-\sqrt5\right\}$$ Below is my attempt.

\begin{align}f(x)&=(x+1)( (x-1) - \sqrt5 ) ( (x-1) + \sqrt5 )\\ &=(x+1)( (x-1)^2 - 5)\\ &=(x+1)(x^2 - 2x - 4)\end{align}

And the answer in the book is $(x+1)(x^2 + 2x -4)$. I'm not sure what's correct?

I'm taking a Math correspondence. The book has very little explanation and few errors. My hunch is that, the answer is wrong. Please advice.


The first root is $-1$ instead of $1$. Sorry about typo.

  • 1
    $\begingroup$ It should be $(x-1)(x^2 - 2x-4)$. $\endgroup$ – Robert Israel May 9 '17 at 15:19
  • $\begingroup$ The book's answer would be for the roots $\{-1, -1+\sqrt{5}, -1-\sqrt{5}\}$, i.e. the negation of the roots. $\endgroup$ – Jaap Scherphuis May 9 '17 at 15:25
  • $\begingroup$ Looks like my calculation is correct. $\endgroup$ – gmail user May 9 '17 at 15:34

When we have the roots $\{a,b,c\}$, we can write $$f(x)=(x-a)(x-b)(x-c)$$We have the roots $\{-1,1+\sqrt 5,1−\sqrt5\}$ so we can write the equation in the following way:

\begin{align}f(x)&=(x-(-1))(x-(1+\sqrt5))(x-(1-\sqrt5))\\ &=(x+1)(x-1-\sqrt5)(x-1+\sqrt5)\\ &=(x+1)(x^2-x+x\sqrt5-x+1-\sqrt5-x\sqrt5+\sqrt5-5)\\ &=(x+1)(x^2-(1-\sqrt5+1+\sqrt5)x+(1+-\sqrt5+\sqrt5-5))\\ &=(x+1)(x^2-2x-4)\\ &=x^3-2x^2-4x+x^2-2x-4\\ &=x^3-x^2-6x-4\end{align}

We can verify this with WolframAlpha and see that it is correct


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