Polynomials

Kushballo7

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Joined
Oct 20, 2005
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17
Show that a polynomial of degree 3 has at most 3 roots and similarly show that a polynomial of degree n has at most n roots.
 
x^3 plus some other stuff can be broken into (x+1)(x+2)(x+3) at the most because there are only 3 x's kind of, because x^3 can only be broken down into 3 x's, if you get what I am saying so therefore x^n can only be broken down into n number of x's

I hope you find this helpful, I'm not too good at explaining stuff, but I like to try, I wasn't sure if you needed an official-y proof, cuz i'm way too lazy to do that for ya
 
You're doing this for calculus, not abstract algebra or some such, so I'm guessing they're wanting you to use something from calculus. :wink:

My guess: A third-degree polynomial is of the form y = ax<sup>3</sup> + bx<sup>2</sup> + cx + d. The derivative then is y' = 3ax<sup>2</sup> + 2bx + c. Setting this equal to zero, you can solve the quadratic equation to show that the derivative has one or two real zeroes. This tells you that the original cubic had at most [how many?] critical points and thus at most [how many?] max/min points.

Cubics must have at least one x-intercept. If there are no critical points, can the curve ever turn back on itself and cross the x-axis again? How many times can the curve turn itself around?

Note: I don't know that this is what your book is looking for. This is just my off-the-cuff reaction.

Eliz.
 
Yeah the proof has to be calculus related. I tried some stuff and Rolle's Theorem seems to be helpful. I just dont know how to use it for an ambiguous case -- to prove that a polynomial of degree n has at most n roots.
 
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