Question

which of the number(s) below are potential roots of the function?

(q(x) = 6x^3 + 19x^2 - 15x - 28)

(pm \frac{2}{3})

(pm \frac{7}{2})

(pm \frac{1}{7})

(pm 6)

(pm 14)

(pm \frac{3}{5})

Explanation:

🆕 New Concept Discovered: Rational Root Theorem
Finding possible rational solutions using factors of the first and last coefficients.

Step 1: Identify the leading coefficient and constant term

To find the potential rational roots of a polynomial function, we use the Rational Root Theorem. This theorem states that any potential rational root must be in the form:
\[ \pm \frac{p}{q} \]
where \( p \) is a factor of the constant term (the last number) and \( q \) is a factor of the leading coefficient (the coefficient of the highest power of \( x \)).

For the function \( q(x) = 6x^3 + 19x^2 - 15x - 28 \):

• The constant term is \( -28 \), so we look for factors of \( 28 \). Let's call these \( p \).
• The leading coefficient is \( 6 \), so we look for factors of \( 6 \). Let's call these \( q \).

Step 2: List the factors of each term

Find all integer factors for both numbers:

• Factors of the constant term \( 28 \) (\( p \)):

\[ p \in \{1, 2, 4, 7, 14, 28\} \]

• Factors of the leading coefficient \( 6 \) (\( q \)):

\[ q \in \{1, 2, 3, 6\} \]

Step 3: Test the given options

We construct potential roots by dividing any factor \( p \) by any factor \( q \):
\[ \text{Potential Roots} = \pm \frac{p}{q} \]

Let's check each of the given choices to see if they can be written in this form:

1. \( \pm \frac{2}{3} \):

• Numerator \( 2 \) is a factor of \( 28 \) (\( p = 2 \)).
• Denominator \( 3 \) is a factor of \( 6 \) (\( q = 3 \)).
• This is a potential root.

1. \( \pm \frac{7}{2} \):

• Numerator \( 7 \) is a factor of \( 28 \) (\( p = 7 \)).
• Denominator \( 2 \) is a factor of \( 6 \) (\( q = 2 \)).
• This is a potential root.

1. \( \pm \frac{1}{7} \):

• Numerator \( 1 \) is a factor of \( 28 \) (\( p = 1 \)).
• Denominator \( 7 \) is not a factor of \( 6 \) (\( q

eq 7 \)).

• This is not a potential root.

1. \( \pm 6 \) (which is \( \pm \frac{6}{1} \)):

• Numerator \( 6 \) is not a factor of \( 28 \) (\( p

eq 6 \)).

• This is not a potential root.

1. \( \pm 14 \) (which is \( \pm \frac{14}{1} \)):

• Numerator \( 14 \) is a factor of \( 28 \) (\( p = 14 \)).
• Denominator \( 1 \) is a factor of \( 6 \) (\( q = 1 \)).
• This is a potential root.

1. \( \pm \frac{3}{5} \):

• Numerator \( 3 \) is not a factor of \( 28 \) (\( p

eq 3 \)).

• Denominator \( 5 \) is not a factor of \( 6 \) (\( q

eq 5 \)).

• This is not a potential root.

Answer:

The potential roots of the function from the given choices are:

• \( \pm \frac{2}{3} \)
• \( \pm \frac{7}{2} \)
• \( \pm 14 \)