The given polynomial is partially visible and reads as:

$x^7 + 2x^6 + \text{(missing terms)} + 6$

The problem asks us to determine which of the given options could not have been an integer root of the polynomial. The possible integer roots of a polynomial are determined by the Rational Root Theorem, which states that any rational root, if it exists, must be a divisor of the constant term (in this case, $6$) divided by the leading coefficient. Since the leading coefficient is 1 (from $x^7$), we are concerned with the divisors of 6.

The divisors of 6 are:

$\pm 1, \pm 2, \pm 3, \pm 6$

Now, let's examine the given options:

• (a) $6$
• (b) $2$
• (c) $4$
• (d) $-3$
• (e) $-2$
• (f) $-1$

Among these, $4$ is not a divisor of 6. Therefore, 4 could not have been an integer root of the polynomial.

The answer is (c) 4.

Would you like further details or have any other questions?

Here are 5 related questions to expand your understanding:

1. What is the Rational Root Theorem and how is it applied in polynomial equations?
2. How do you find all possible rational roots of a polynomial with a non-unit leading coefficient?
3. How do we determine the multiplicity of a root in a polynomial equation?
4. Can you provide an example of how synthetic division is used to test for polynomial roots?
5. How does the degree of a polynomial affect the number of real roots it can have?

Tip: Always check the constant term and leading coefficient when using the Rational Root Theorem!