Question

Find all roots of x^5 + x^4 + x^3 + x^2 + x +1 = 0

Answer 1

1) Degree is odd. There MUST be one Real root.
2) All coefficients are positive. There is NO positive Real root.
3) There may be as many as 5 Negative Real Roots. (I rather doubt there are that many.)
4) If there is to be a RATIONAL Real root, then x = -1 is the ONLY candidate. Sure enough, it works.
5) This makes (synthetic division) \(f(x) = (x+1)(x^{4} + x^{2} + 1)\)
6) Well, isn't that special. This can no be solved completely using the Quadratic Formula.

\(x^{2} = \dfrac{-1\pm\sqrt{-3}}{2}\) -- Well, there are no more Real roots, but we can name all the complex roots with relative ease.

This is a wonderful exploration.

Answer 2

It may be a little simpler to rewrite the expression a bit. The first (and Real) root rather pops out.

\(f(x) = \dfrac{1-x^{6}}{1-x}\) except that \(x \ne 1\)

Answer 3

Thank you so much for all of your explanations. I really didn't even know where to begin. This helps it much more easier to see. Looking at the second equation the first and real root are not popping out at me. I see that you should get 1- x^5 or possibly just x^5 but I know that isn't right.

Answer 4

or is it -5 and 5

Answer 5

You should not get 1-x^5. If you do the division, you should get the original expression for f(x). Just look at that little numerator.

There is nothing here about x = 5 or x = -5.

If \(1-x^{6} = 0\), how many real numbers can you think of to solve that simple equation. Hopefully, you said x = 1 and x = -1. Since x = 1 is NOT in the Domain, x = -1 is all that is left. You can factor it as a difference of squares if you like. \(1 - x^{6} = (1-x^{3})(1+x^{3})\). Is it getting more clear, yet?

Factoring the whole thing:

\(\dfrac{1-x^{6}}{1-x} = \dfrac{(1-x^{3})(1+x^{3})}{1-x} = \dfrac{\left[(1-x)(1+x+x^{2})\right]\left[(1+x)(1-x+x^{2})\right]}{1-x}\)

And simplify: \((1+x+x^{2})\left[(1+x)(1-x+x^{2})\right]\)

That looks a WHOLE LOT like the laborious division and quadratic formula we did above.

Answer 6

Okay. That makes much more sense!! so the simplified answer would be the roots of the problem correct since there is not true value for x

Answer 7

What? That doesn't mean anything. If you can factor it, then factor it. In this case, there were two quite reasonable ways to factor. Then, find all the roots.

1 + x = 0 -- Straight forward.
1 + x + x^2 = 0 -- Requires Quadratic Formula
1 - x + x^2 = 0 -- Requires Quadratic Formula

You will find all five in this way.

Answer 8

Okay. This makes sense . I solved the other ones and got three solutions being 1, -1 + root 5/2, and -1 -root 5/2. The other equation presents non real solutions as you get a negative under the radical.

Answer 9

How did you get those? There is no correct answer in that list.

x = -1 is the only REAL solution.
The other four are Complex. You should have an "i" in there, somewhere.

Answer 10

Oh okay. It is what I got using the quadratic formula. This stuff I really struggle in. I greatly appreciate all of your help!