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A polynomial function $g(x)$ with integer coefficients has a leading coefficient of $5$ and a constant term of $-2$ . According to the Rational Root Theorem, which of the following are possible roots of $g(x)$ ? $\newline$ Multi-select Choices: $\newline$ (A) $-2$ $\newline$ (B) $2$ $\newline$ (C) $\frac{1}{5}$ $\newline$ (D) $\frac{5}{13}$

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Rational root theorem

Full solution

Q. A polynomial function

g(x)

with integer coefficients has a leading coefficient of

5

and a constant term of

-2

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

g(x)

?

\newline

Multi-select Choices:

\newline

(A)

-2

\newline

(B)

2

\newline

(C)

\frac{1}{5}

\newline

(D)

\frac{5}{13}

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), of a polynomial equation with integer coefficients must be such that $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient.
Constant Term Factors: List the factors of the constant term $-2$ : $-1$ , $1$ , $-2$ , $2$ .
Leading Coefficient Factors: List the factors of the leading coefficient $5$ : $-1$ , $1$ , $-5$ , $5$ .
Possible Roots Calculation: According to the Rational Root Theorem, the possible rational roots are the combinations of the factors of the constant term over the factors of the leading coefficient. So, we get the possible roots as: $-\frac{1}{1}$ , $\frac{1}{1}$ , $-\frac{2}{1}$ , $\frac{2}{1}$ , $-\frac{1}{5}$ , $\frac{1}{5}$ , $-\frac{2}{5}$ , $\frac{2}{5}$ , $-\frac{1}{-1}$ , $\frac{1}{-1}$ , $\frac{1}{1}$ $0$ , $\frac{1}{1}$ $1$ , $\frac{1}{1}$ $2$ , $\frac{1}{1}$ $3$ , $\frac{1}{1}$ $4$ , $\frac{1}{1}$ $5$ .
Simplify Possible Roots: Simplify the list of possible roots to get unique values: $-1$ , $1$ , $-2$ , $2$ , $-\frac{1}{5}$ , $\frac{1}{5}$ , $-\frac{2}{5}$ , $\frac{2}{5}$ .
Check Given Choices: Now, we check which of these possible roots are in the given choices. We have $-2$ (A), $2$ (B), and rac{1}{5} (C) as part of our list. rac{5}{13} (D) is not a possible root because $13$ is not a factor of the leading coefficient $5$ .

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