Find a degree 6 polynomial with - 8 as a root, no other roots, and in which the coefficient of x^6 is 16. Assume that all (non-constant) factors of the polynomial correspond to real zeroes.
The question seems ambiguous. I suggest you repost with the exact original wording.
A polynomial of even degree cannot have one single real root. It is possible to have two coincident roots, but not just one.
It then goes on to say that
"Assume that all (non-constant) factors of the polynomial correspond to real zeroes."
So is the question asking for "a single root", or 6?
To find a degree 6 polynomial with -8 as a root and no other roots, we need to factorize the polynomial. Since all non-constant factors of the polynomial correspond to real zeroes, this means we also need to ensure that -8 does not repeat as a root.
Let's start by writing the polynomial in factored form:
P(x) = (x - (-8))(a)(b)(c)(d)(e)(f)
where a, b, c, d, e, and f are constants.
To ensure that -8 does not repeat as a root, we need to make sure that (x - (-8)) is not part of the factored form. In other words:
(x - (-8)) = (x + 8) should not be one of the factors.
Now, let's also assume that the coefficient of x^6 is 16. We can express the polynomial as:
P(x) = 16(x + 8)(a)(b)(c)(d)(e)(f)
To meet the conditions, we need to select the values of a, b, c, d, e, and f such that P(x) is a degree 6 polynomial and -8 is the only root.
Since we have 6 variables to choose from, we can simply assign values to each variable:
P(x) = 16(x + 8)(1)(1)(1)(1)(1)(1)
P(x) = 16(x + 8)
Therefore, one possible degree 6 polynomial that satisfies the given conditions is P(x) = 16(x + 8).