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

In summary: So if the given equation has roots in y, then the equation has roots in x as well.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.
  • #1
rock.freak667
Homework Helper
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Homework Statement



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

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

Homework Equations



[tex]\sum \alpha=\frac{-b}{a}

\

\sum \alpha\beta=\frac{c}{a}

\
\sum \alpha\beta\gamma=\frac{-d}{a}[/tex]

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?
 
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  • #2
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.
 
  • #3
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 [itex]\alpha[/itex], [itex]\beta[/itex], and [itex]\gamma[/itex] make the first factor 0, they make the entire polynomial 0.
 
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  • #4
HallsofIvy said:
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 [itex]\alpha[/itex], [itex]\beta[/itex], and [itex]\gamma[/itex] 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.
 
  • #5
symbolipoint said:
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?
 
  • #6
symbolipoint said:
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!
 
  • #7
HallsofIvy said:
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.
 
  • #8
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 6th degree polynomial (after substituting x2 for y) has the first 3rd degree polynomial equations as a factor.
 

1. What is the definition of a polynomial equation?

A polynomial equation is an algebraic equation that contains one or more variables and consists of terms that are formed by multiplying a constant number with a variable raised to a non-negative integer power. The highest power of the variable in a polynomial equation is known as the degree of the polynomial.

2. How do you solve a polynomial equation?

To solve a polynomial equation, you need to follow a step-by-step process that involves simplifying the equation, factoring it, and finding the roots or solutions. First, simplify the equation by combining like terms. Then, factor the polynomial using techniques such as grouping, difference of squares, or the quadratic formula. Finally, solve for the unknown variable by setting the factors equal to zero and finding the corresponding values.

3. What is the difference between a root and a solution of a polynomial equation?

A root of a polynomial equation is a value that makes the equation equal to zero, while a solution is any value that makes the equation true. All roots are solutions, but not all solutions are roots. This is because a polynomial equation can have complex or imaginary solutions that do not make the equation equal to zero.

4. How do you find the value of Sn in a polynomial equation?

The value of Sn, also known as the sum of the roots, can be found by using Vieta's formulas. These formulas state that the sum of the roots of a polynomial equation is equal to the opposite of the coefficient of the second-to-highest degree term divided by the coefficient of the highest degree term. You can also find the value of Sn by adding the individual roots together.

5. Can you use the roots of a polynomial equation to find the equation itself?

Yes, you can use the roots of a polynomial equation to find the equation itself. This process is known as reverse factoring or reconstruction. To do this, you need to know the roots and at least one point on the polynomial. Then, you can use the roots to create the factors and use the given point to determine the leading coefficient. Finally, multiply all the factors together to get the original polynomial equation.

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