Finding number of zeroes in a polynomial?

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Discussion Overview

The discussion revolves around determining the number of zeroes in a polynomial equation, specifically examining the polynomial f(x) = 2x + 3 * (3x^2 + 3) - x^2 + 5. Participants explore the degree of the polynomial and the implications for the number of roots, including real and complex roots.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant claims the polynomial is a 3rd-degree polynomial and asks how many zeroes it has.
  • Another participant explains that an nth degree polynomial has at most n zeroes, suggesting a general form for cubic polynomials.
  • Some participants challenge the degree of the polynomial, asserting that the original equation does not represent a cubic polynomial.
  • There is a discussion about the nature of roots, with some stating that a cubic polynomial can have 1, 2, or 3 real roots, while others mention the possibility of complex roots.
  • One participant provides an expanded form of the polynomial, arguing that it is actually a quadratic rather than cubic.
  • Another participant suggests that the original poster may have intended a different expression that would indeed be cubic.
  • Participants discuss the concept of root multiplicity, clarifying that a cubic polynomial can have repeated roots.

Areas of Agreement / Disagreement

Participants do not reach consensus on the degree of the polynomial or the number of zeroes it has. Multiple competing views remain regarding the interpretation of the polynomial and the implications for its roots.

Contextual Notes

There are unresolved issues regarding the correct interpretation of the polynomial expression, including potential missing parentheses and the resulting degree of the polynomial. The discussion also highlights the complexity of counting roots, particularly in relation to multiplicity.

moonman239
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Let's say I have the equation f(x) = 2x + 3 * (3x^2 + 3) - x^2 + 5. If my algebra is right, this is a 3rd-degree polynomial. How many zeroes does this equation have? How did you figure that out?
 
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if you remember: an nth degree polynomial has n factors of the form f(x) = (x-a)*(x-b)*...
hence it has at most n zeros. ( I assumed the factors for x were 1)

So in your case for a 3rd degree:

f(x) = (x -a) * (x - b ) * ( x - c )
___ = (x^2 - (a+b)x + ab) * (x - c)
___ = x^3 - (a+b+c)x^2 + (ab+ac+bc)x + abc
 
Last edited:
Am I missing something?? The equation in the OP does not have third degree...
 
you're right, i was addressing find the number of roots only.
 
micromass said:
Am I missing something?? The equation in the OP does not have third degree...

2x2 + 3 + (3X2 + 3) = 6x3 + 6x + 6x2 + 9. That's why I thought the degree was 3.
 
moonman239 said:
2x2 + 3 + (3X2 + 3) = 6x3 + 6x + 6x2 + 9.

I have no clue why you think this equality is true.
 
I think you meant (2x+3)*(3x^2+3) right?
 
Assuming it is a cubic, there will be 1, 2 or 3 real roots. There are always 3 roots altogether, but some may be complex pairs. That can reduce it to 1 real root. There is also the borderline case where two real roots coincide, making only 2 values.
 
moonman239 said:
Let's say I have the equation f(x) = 2x + 3 * (3x^2 + 3) - x^2 + 5. If my algebra is right, this is a 3rd-degree polynomial.
Based on what you wrote, your algebra is incorrect. Expanding what you wrote, I get
f(x) = 2x + 9x2 + 9 - x2 + 5
= 8x2 + 2x + 14, which is NOT a cubic.

I suspect that you are missing some parentheses, and actually meant
f(x) = (2x + 3)*(3x2 + 3) - x2 + 5, which IS a cubic.
moonman239 said:
How many zeroes does this equation have? How did you figure that out?
 
  • #10
haruspex said:
Assuming it is a cubic, there will be 1, 2 or 3 real roots. There are always 3 roots altogether, but some may be complex pairs. That can reduce it to 1 real root. There is also the borderline case where two real roots coincide, making only 2 values.

It can't have two real roots.
 
  • #11
It can have three real roots, two of which are the same, having two distinct real roots. That is what haruspex was talking about. He was not counting "multiplicity".

For example, x^3- x^2= 0 has two distinct roots- 0 and 1. 0 is a double root.
 
Last edited by a moderator:
  • #12
HallsofIvy said:
It can have three real roots, two of which are the same, having two distinct real roots. That is what haruspex was talking about. He was not counting "multiplicity".

For example, x^3- x^2= 0 has two distinct roots- 0 and 1. 0 is a double root.

YEs, I see now. I just re-read his post.
 
Last edited by a moderator:

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