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

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

micromass

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

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.

Robert1986

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, [itex]x^3- x^2= 0[/itex] has two distinct roots- 0 and 1. 0 is a double root.

Robert1986

YEs, I see now. I just re-read his post.