Finding number of zeroes in a polynomial?

[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. How many zeroes does this equation have? How did you figure that out?

 

[jedishrfu]

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

 

[micromass]

Am I missing something?? The equation in the OP does not have third degree...

 

[jedishrfu]

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

 

[moonman239]

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.

 

[micromass]

2x² + 3 + (3X² + 3) = 6x³ + 6x + 6x² + 9.

I have no clue why you think this equality is true.

 

[jedishrfu]

I think you meant (2x+3)*(3x^2+3) right?

 

[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.

 

[Mark44]

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.

How many zeroes does this equation have? How did you figure that out?

 

[Robert1986]

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.

 

[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.

 

[Robert1986]

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.