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A polynomial function $f(x)$ with integer coefficients has a leading coefficient of $1$ and a constant term of $-16$ . According to the Rational Root Theorem, which of the following are possible roots of $f(x)$ ? $\newline$ Multi-select Choices: $\newline$ (A) $-1$ $\newline$ (B) $-14$ $\newline$ (C) $-\frac{4}{5}$ $\newline$ (D) $-3$

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Rational root theorem

Full solution

Q. A polynomial function

f(x)

with integer coefficients has a leading coefficient of

1

and a constant term of

-16

. According to the Rational Root Theorem, which of the following are possible roots of

f(x)

?

\newline

Multi-select Choices:

\newline

(A)

-1

\newline

(B)

-14

\newline

(C)

-\frac{4}{5}

\newline

(D)

-3

Rational Root Theorem: The Rational Root Theorem states that any rational root, in the form of $\frac{p}{q}$ (where $p$ and $q$ are integers and $q$ is not zero), of a polynomial equation with integer coefficients must be a factor of the constant term divided by a factor of the leading coefficient.
Factors of Constant Term: Since the leading coefficient is $1$ , any rational root must be a factor of the constant term $-16$ . The factors of $-16$ are $\pm 1$ , $\pm 2$ , $\pm 4$ , $\pm 8$ , and $\pm 16$ .
Checking Options: Check each of the given options against the factors of $-16$ . $\newline$ (A) $-1$ is a factor of $-16$ , so it could be a root. $\newline$ (B) $-14$ is not a factor of $-16$ , so it cannot be a root. $\newline$ (C) $-\frac{4}{5}$ is not an integer and thus cannot be a root since the coefficients are integers. $\newline$ (D) $-3$ is not a factor of $-16$ , so it cannot be a root.

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