Roots of a polynomial of degree 4

So x+ x-1 is a root of Q is equivalent to p(x)= 0. In summary, we are trying to prove that a complex number x satisfies the equation p(x) = 0 if and only if the number s = x + x^-1 is a root of the polynomial Q(s) = s^2 + as + (b-2). This can be shown by observing that Q(x + x^-1) = p(x)/x^2 and noting that for a fraction to equal 0, the numerator must also be 0. Therefore, x + x^-1 is a root of Q if and only if p(x) = 0.
  • #1
Mathman23
254
0
(*)[tex]p(x) = x^4 + ax^3 + bx^ 2 + ax + 1 = 0[/tex]

where [tex]a,b \in \mathbb{C}[/tex]

I would like to prove that a complex number x makes (*) true iff

[tex]s = x + x^{-1}[/tex] is a root of the [tex]Q(s) = s^2 + as + (b-2) [/tex]

I see that that [tex]Q(x + x^{-1}) = \frac{p(x)}{x^2}[/tex]

Then to prove the above do I then show that p(x) and Q((x + ^{-1}) shares roots?

Sincerely Yours
MM23
 
Physics news on Phys.org
  • #2
Yes. Note that Q(s) and p(x) must have coincident roots.
 
  • #3
Mathman23 said:
(*)[tex]p(x) = x^4 + ax^3 + bx^ 2 + ax + 1 = 0[/tex]

where [tex]a,b \in \mathbb{C}[/tex]

I would like to prove that a complex number x makes (*) true iff

[tex]s = x + x^{-1}[/tex] is a root of the [tex]Q(s) = s^2 + as + (b-2) [/tex]

I see that that [tex]Q(x + x^{-1}) = \frac{p(x)}{x^2}[/tex]

Then to prove the above do I then show that p(x) and Q((x + ^{-1}) shares roots?

Sincerely Yours
MM23
The onlly way a fraction can be 0 is if the numerator is 0. Isn't it obvious from [tex]Q(x + x^{-1}) = \frac{p(x)}{x^2}[/tex]
that Q(x+ x-1)= 0 if and only if p(x)= 0?
 

Related to Roots of a polynomial of degree 4

1. What is a polynomial of degree 4?

A polynomial of degree 4 is an algebraic expression that contains a variable raised to the 4th power and has coefficients and constants. It can be written in the form ax^4 + bx^3 + cx^2 + dx + e, where a, b, c, d, and e are constants and x is the variable.

2. What are the roots of a polynomial of degree 4?

The roots of a polynomial of degree 4 are the values of the variable that make the polynomial equal to zero. In other words, they are the solutions to the equation ax^4 + bx^3 + cx^2 + dx + e = 0.

3. How do you find the roots of a polynomial of degree 4?

To find the roots of a polynomial of degree 4, you can use the rational root theorem, synthetic division, or the quadratic formula. You can also use a graphing calculator to approximate the roots.

4. Can a polynomial of degree 4 have complex roots?

Yes, a polynomial of degree 4 can have complex roots. This means that the solutions to the equation ax^4 + bx^3 + cx^2 + dx + e = 0 can be complex numbers, which are numbers with a real and imaginary part. Complex roots usually come in pairs of complex conjugates.

5. How many roots does a polynomial of degree 4 have?

A polynomial of degree 4 can have up to 4 roots, but it is not guaranteed to have 4 distinct roots. It may have repeated roots or complex roots. The fundamental theorem of algebra states that a polynomial of degree n has n complex roots, counting multiplicity.

Similar threads

  • Calculus and Beyond Homework Help
Replies
18
Views
2K
  • Calculus and Beyond Homework Help
Replies
24
Views
892
  • Calculus and Beyond Homework Help
Replies
7
Views
598
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
750
  • Calculus and Beyond Homework Help
Replies
6
Views
257
  • Calculus and Beyond Homework Help
Replies
5
Views
667
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
722
Back
Top