## Finding number of zeroes in a polynomial?

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?

 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
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Am I missing something?? The equation in the OP does not have third degree...

## Finding number of zeroes in a polynomial?

you're right, i was addressing find the number of roots only.

 Quote by micromass 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.

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 Quote by moonman239 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?
 Recognitions: Homework Help Science Advisor 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.

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 Quote by moonman239 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.
 Quote by moonman239 How many zeroes does this equation have? How did you figure that out?

 Quote by haruspex 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.

 Recognitions: Gold Member Science Advisor Staff Emeritus 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.

 Quote by HallsofIvy 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.