# Find the number of distinct real roots of a polynomial

Suppose we have the polynomial

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

where a-e are real.

$$f(x)= 1$$ 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?

• 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 ? – Nicolas FRANCOIS Oct 22 '18 at 12:42
• How many distinct real roots does $x^2(x-1)(x-2)=0$ have? – Mark Bennet Oct 22 '18 at 12:43
• 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... – DonAntonio Oct 22 '18 at 12:47
• @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. – darren86 Oct 22 '18 at 12:52
• 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. – Nicolas FRANCOIS Oct 22 '18 at 12:54