Give an example and explain why a polynomial can have fewer x-intercepts than its number of roots.

A polynomial can have fewer x-intercepts than its number of roots when a pair of complex conjugate roots exist.

Example:
Consider the 4-th degree polynomial
f(x) = x⁴ - x³ - x² - x - 2

According to the Remainder theorem
f(-1) = 1 + 1 - 1 + 1 - 2 = 0
Therefore (x + 1) is a factor.

f(2) = 16 - 8 - 4 - 2 - 2 = 0
Therefore (x - 2) is a factor.

(x+1)*(x-2) = x² - 2x + x - 2 = x² - x - 2
To find the remaining factor, perform long division.
                            x²      + 1
          -------------------------------
x²-x-2 | x⁴  - x³  -  x²  - x  - 2
             x⁴  - x³ - 2x²
            -----------------------------
                             x² - x - 2
                             x² - x - 2
Therefore
f(x) = (x+1)(x-2)(x²+1)

Notice that (x² + 1) has no real factors.
However,
x² + 1 = (x + i)(x - i),
so it has a pair of conjugate zeros +i and -i.

A graph of the function confirms that there are only two real zeros (shown in red color).

Are there any statements?

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