Suppose we have the polynomial

$f(x) = a x^4 + bx^3 + cx^2+dx+e $

where a-e are real.

Can someone please help explain why the following is definitely not true:

$f(x)= 1$ has 1 distinct real solution

$f(x)= 2$ has 3 distinct real solutions

$f(x)= 3$ has 2 distinct real solutions

and $f(x)= 4$ has 4 distinct real solutions

yet the following could be true

$f(x)= 1$ has 1 distinct real solution

$f(x)= 2$ has 2 distinct real solutions

$f(x)= 3$ has 4 distinct real solutions

and $f(x)= 4$ has 3 distinct real solutions

I thought that by the fundamental theorem of algebra that f(x) has 4 roots. Therefore $f(x) - \alpha$ for some constant $\alpha$ should have four roots. And because complex roots come in pairs how could for example $f(x)= 4$ have 3 distinct real solutions?

  • $\begingroup$ FTA gives you the number of complex roots. Here, you ask about real roots. Do you want a polynomial that verifies ALL the conditions, or one at a time ? $\endgroup$ – Nicolas FRANCOIS Oct 22 '18 at 12:42
  • $\begingroup$ How many distinct real roots does $x^2(x-1)(x-2)=0$ have? $\endgroup$ – Mark Bennet Oct 22 '18 at 12:43
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    $\begingroup$ For example: $\;f(x)=x^4+2\;$ shows that 1-2 could be false (meaning: not necessarily true always, which I believe is what you mean) . Also, $\;f(x)=x^4+4\;$ shows the last one could be false. The difference between both sets of questions: "definitely not true" (??), and "could be true" seems to be pretty foggy... $\endgroup$ – DonAntonio Oct 22 '18 at 12:47
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    $\begingroup$ @DonAntonio the question states that there is no polynomial satisfies the first set on conditions, but you could find a polynomial which satisfies the second set. $\endgroup$ – darren86 Oct 22 '18 at 12:52
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    $\begingroup$ I think the best way is to draw a picture. An odd number of roots (for $P(x)-\alpha$) means there is a multiple root, hence an horizontal tangent. I'll give you an example that satisfies the second set of solutions, you'll see what I mean. Anyway, this is a real analysis problem, not an algebra problem. $\endgroup$ – Nicolas FRANCOIS Oct 22 '18 at 12:54

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