Find all poss. real n so roots of 3x^3 + 11x^2 + 8x + n = 0 are non-real

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popcorn123

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Determine all possible values of n, NER. such that the equation <> has no real roots

<> = 3x^3 + 11x^2 + 8x + n =0
 
popcorn123 said:
Determine all possible values of n, NER. such that the equation <> has no real roots

<> = 3x^3 + 11x^2 + 8x + n =0
Click to expand...
What have you tried? How far have you gotten? Where are you stuck? Please be complete. Thank you! ;)
 
Ishuda said:
Jut out of curiosity, would you please tell us where this problem came from?
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It's actually from our homework: our teacher gets it from the University of Waterloo website. I can give you the link:

But the problem is these questions - I don't even know where to start. I try to apply all we've done - Factor theorem, synthetic division, etc - but seems there is something here I'm not seeing.

If you can help me that would be much appreciated. I've tried them a million times and failed - that's why Im here.


Here;s the link: http://courseware.cemc.uwaterloo.ca/8?gid=146

There is different categorizes - this one is from Solving Polynomial Equations number 12. Make sure you click the exercise category once you're there. Thanks
 
A cubic polynomial, with real coefficients, has at least one real root.
 
popcorn123 said:
It's actually from our homework: our teacher gets it from the University of Waterloo website. I can give you the link:

But the problem is these questions - I don't even know where to start. I try to apply all we've done - Factor theorem, synthetic division, etc - but seems there is something here I'm not seeing.

If you can help me that would be much appreciated. I've tried them a million times and failed - that's why Im here.


Here;s the link: http://courseware.cemc.uwaterloo.ca/8?gid=146

There is different categorizes - this one is from Solving Polynomial Equations number 12. Make sure you click the exercise category once you're there. Thanks
Click to expand...
As Subhotosh Khan points out, a cubic polynomial with real coefficients has at least one root so there is no n for which there is no real roots to
f(x;n) = 3x^3 + 11x^2 + 8x + n
That was the main reason I asked the question.

The easiest way to see this, IMO, is to use the intermediate value theorem.
 
popcorn123 said:
Determine all possible values of n, NER. such that the equation <> has no real roots

<> = 3x^3 + 11x^2 + 8x + n =0...

Here;s the link: http://courseware.cemc.uwaterloo.ca/8?gid=146
Click to expand...
Um... This question might be easier if one uses the correct text of the original exercise (#12 at the bottom of this page):

\(\displaystyle \mbox{Determine all possible values of }\, n,\, n\, \in\, \mathbb{R},\, \mbox{ such that the equation }\)

. . . . .\(\displaystyle 3x^3\, +\, 11x^2\, +\, 8x\, +\, n\, =\, 0\, \)\(\displaystyle \color{red}{\mbox{ has two equal real roots.}}\)
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With this information, you know that you must have two identical linear factors. Does this help at all? ;)
 
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More specifically:

1. The fundamental theorem of polynomials is that every polynomial of degree "n" has exactly "n" roots, some of which may be nonreal and/or repeated.

2. If the polynomial has all real coefficients, then any nonreal solutions appear as complex conjugates - i.e. in PAIRS.

So that means that any polynomial of odd degree has to have at least one real solution, as any nonreal solutions will have to have a conjugate pair.
 
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