Fifth grad equation. No computers solutions, don´t suggest synthetic division?

Question

If you know an algebraic method to solve this one, I'd like to know how you do it.

I got 2 as a root but I am stucked in the 4th grad quotient of dividing the equation by x-2

x^5 + 2x^4 - 11 x^3 - 22 x^2 + 24 x + 64 = 0

Please, avoid general remarks like: "Use synthetic division" or "Use the rational roots theorem" (I already checked all of them and there are none) or "Take derivatives and see what the extrems are". I already did this too and I already know in which interval the roots more or less are, I found them in an approximated way too.

The thing is that this questions is in an exam I have to solve and I'd like to know how the equation is supposed to be solved, and if there is a algebraic way to do this. And no, computers aren't available during the exam.

Thanks

I don't have anything wrong with my procedure. The equation is this way, it's in the proposal. Solve ... (the equation here)

But thanks, your method perhaps works better than the things I tried.

Ana

Thanks, Lobosito, I will check this later.

And thanks to you too, Starwhitedwarf. I think that perhaps this equation has a root like this a + bVp and its conjugated since this is the last part of a problem where you had to prove that, if, there were one root like that, the conjugate binomial was a root too. But I tried to use this fact and, as you said, I got too nasty equations. I drew the function and I drew f(x) -16 (I already know what the roots are since the students were supposed to solve it in a previous part), and I noticed that there were only 2 roots. So, I used this:

(x- a - bVp)(x- a + bVp) = (x - a)^2 - b^2p

I multiplied this by a quadratic factor, x^2 + mx + n. But, as you say, the equations are too nasty.

Ok, I think I've got the answer to this problem. I just read the proposal again and this equation hast to be factored in Q. So, I guess that, what the students are supposed to do, it just to prove that there are no other rational roots and to express the equation as a product of
(x-2) by the quartic quotient

Anyway, I wrote to the professor that sent me this proposal and I will let you (I mean the ones of you I can email) know what she replies.

Sorry I bothered all of you.

Thanks, Steiner.

Answer 1

Hi, Ana!
What kind of an exam problem is this??
Sounds like some sadistic examiner proposed it.
I tried to solve it algebraically and only got
as far as the resultant cubic. When I plugged
that into Cardan's formulas I got something
terribly messy. Yes, I used PARI to help with the
calculations.
So let me show you what I have and see what
we can do from there. Probably the resultant
cubic will have to be done by iteration.
Your equation is
x^5 + 2x^4 - 11 x^3 - 22 x^2 + 24 x + 64 = 0
or
(x-2)(x^4 + 4x³ -3x² -28x - 32) = 0
So x = 2 is a root
and we have to solve
x^4 + 4x³ -3x² -28x - 32 = 0. (*)
A first step in solving is to "reduce" the quartic, i.e.,
get rid of the cubed term
To do this, substitute x = u-1 into (*). We get
u^4 -9u² -14u-10 = 0. (**)
(This is an example of a Tschirnhaus transformation)
If we follow the method of solving the quartic
in http://mathworld.wolfram.com/QuarticEquation.html
we find that the resolvent cubic for this equation is
t³ + 9t² + 40t + 164 = 0.
We must now find a root of this cubic to
solve (**).
Let s = t - 3
and we get
s³ + 13s + 98 = 0.
But this has no rational roots and Cardan's
formulas give an extremely messy expression for t.
I suggest solving the cubic by Newton's method
and then using the approximate solution to solve (**).

Answer 2

Very nasty equation to appear in an exam.
This equation does not have rational solutions other than two,
so I cannot see other elegant ways of solving except to use iteration like newton's method.
If we try writing the quartic as the product of 2 quadratics with coefficients to be determined and use the quadratic formula on the resulting quasratics, we will have some nasty equations to try and solve, so I do think newton's method is the best way to go about it.
Sorry, I do not have a better answer.

EDIT: yes anna, it will surely have roots of this form because for any polynomial we can always go down to product of linears and quadratics.

Answer 3

use newton raphson method
x=x-f(x)/f'(x))=

x=x-[x^5 + 2x^4 - 11 x^3 - 22 x^2 + 24 x + 64]/[5x^4+8x^3-33x^2-44x+24].
plug in the value of x in your calculator and you would be able to compute a new value of x and plug it again and again until it converges to a certain value..

If you hae known at least one correct root of f(x), then you can use the synthetic division, may be you have just gone wrong with yor procedures.

Answer 4

x=2. Solve it empirically.

Answer 5

i will give a solution tomorrow