Discover the Roots of Polynomials: Solving Equations and Finding Values of S_n

[rock.freak667]

Homework Statement

The roots of the equation $x^3-x-1=0$ are $\alpha,\beta,\gamma$
$S_n=\alpha^n +\beta^n +\gamma^n$

(i)Use the relation y=x$^2$ to show that $\alpha^2,\beta^2,\gamma^2$
are roots of the equation
$y^3-2y^2+y-1=0$
(ii)Hence, or otherwise find the value of $S_4$
(iii)Find $S_8,S_{12},S_{16}$

Homework Equations

$$\sum \alpha=\frac{-b}{a}<br /> <br /> \<br /> <br /> \sum \alpha\beta=\frac{c}{a}<br /> <br /> \<br /> \sum \alpha\beta\gamma=\frac{-d}{a}$$

The Attempt at a Solution

Just need help with the first part for now. Using I substituted y=x^2 into the equation they gave me in hopes to get back the original equation but that did not work out. Should I just find the sum of the roots and then the sum of the squares of the roots for the original equation and then find the sum of the roots for the eq'n in y and show that they are equal?

 

[symbolipoint]

Are you permitted to use rational roots theorem, and fundamental theorem of algebra? The way you ask the question gives the impression that you are studying something beyond PreCalculus.

 

[HallsofIvy]

You don't need anything deep like that. If y= x², then it is easy to show that y³- 2y²+ y- 1= x⁶- 2x⁴+ x²- 1= (x³-x-1)(x³-x+ 1). Since $\alpha$, $\beta$, and $\gamma$ make the first factor 0, they make the entire polynomial 0.

 

[rock.freak667]

You don't need anything deep like that. If y= x², then it is easy to show that y³- 2y²+ y- 1= x⁶- 2x⁴l+ x²- 1= (x³-x-1)(x³-x+ 1). Since $\alpha$, $\beta$, and $\gamma$ make the first factor 0, they make the entire polynomial 0.

oh thanks. No wonder I was confused. Shall attempt the other parts in a while.

 

[symbolipoint]

Are you permitted to use rational roots theorem, and fundamental theorem of algebra? The way you ask the question gives the impression that you are studying something beyond PreCalculus.

I later did try what I suggested and found that the typical methods of "College Algebra" did not give ready/quick solutions. I found a real zero of about 1.3244... something, and two complex zeros. I have never before seen that alpha, beta, gamma stuff. Again, what course is that from? Is it something beyond PreCalculus?

 

[HallsofIvy]

I later did try what I suggested and found that the typical methods of "College Algebra" did not give ready/quick solutions. I found a real zero of about 1.3244... something, and two complex zeros. I have never before seen that alpha, beta, gamma stuff. Again, what course is that from? Is it something beyond PreCalculus?

You would use "rational roots" theorem and "Fundamental Theorem of Algebra" to try to FIND roots to the equation. Here you are already given the roots and want to show that they are roots. It's not as complicated as you seem to think!

 

[symbolipoint]

You would use "rational roots" theorem and "Fundamental Theorem of Algebra" to try to FIND roots to the equation. Here you are already given the roots and want to show that they are roots. It's not as complicated as you seem to think!

I tried synthetic division for possible roots of +1 and -1, and they gave remainders. I then resorted to a graphing calculator and found a real root (not certain if it is rational or irrational) of about 1.3224. Using this value and again synthetic division, I came to two complex zeros or in any case, just resulting quadratic factor.

I could not pick any better rational values to test. According to the textbook of Precalculus, we pick using factors of the constant term and coefficient of the leading term, the positive and the negative values. These are based on integers.

 

[HallsofIvy]

Apparently you simply did not understand what the question was asking! It is obvious, from the "rational root theorem" you cite, that neither of these equations has rational roots! The problem did not ask us to find any rational roots, or find any roots at all. It only asked us to show that roots of one equation must satisfy the other. That true because the given 6^th degree polynomial (after substituting x² for y) has the first 3^rd degree polynomial equations as a factor.