Question

according to the rational root theorem, (-\frac{2}{5}) is a potential rational root of which function?

(f(x) = 4x^4 - 7x^2 + x + 25)
(f(x) = 9x^4 - 7x^2 + x + 10)
(f(x) = 10x^4 - 7x^2 + x + 9)
(f(x) = 25x^4 - 7x^2 + x + 4)

Explanation:

State the Rational Root Theorem

Using the Rational Root Theorem knowledge point
For a polynomial function with integer coefficients:

$$ f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 $$

Any rational root \(x = \frac{p}{q}\) (in lowest terms) must satisfy:

• \(p\) is a factor of the constant term \(a_0\).
• \(q\) is a factor of the leading coefficient \(a_n\).

Analyze the given potential root

Using the Rational Root Theorem knowledge point
We are given the potential rational root:

$$ x = -\frac{2}{5} $$

Here, the numerator is \(p = \pm 2\) and the denominator is \(q = \pm 5\).
Therefore, for a polynomial to have \(-\frac{2}{5}\) as a potential rational root:

• The constant term \(a_0\) must be divisible by \(2\).
• The leading coefficient \(a_n\) must be divisible by \(5\).

Evaluate the given polynomial options

Using the Rational Root Theorem knowledge point
Let's test each option:

1. \(f(x) = 4x^4 - 7x^2 + x + 25\)

• Leading coefficient \(a_n = 4\) (not divisible by \(5\))
• Constant term \(a_0 = 25\) (not divisible by \(2\))

1. \(f(x) = 9x^4 - 7x^2 + x + 10\)

• Leading coefficient \(a_n = 9\) (not divisible by \(5\))
• Constant term \(a_0 = 10\) (divisible by \(2\))

1. \(f(x) = 10x^4 - 7x^2 + x + 9\)

• Leading coefficient \(a_n = 10\) (divisible by \(5\))
• Constant term \(a_0 = 9\) (not divisible by \(2\))

1. \(f(x) = 25x^4 - 7x^2 + x + 4\)

• Leading coefficient \(a_n = 25\) (divisible by \(5\), since \(5\) is a factor of \(25\))
• Constant term \(a_0 = 4\) (divisible by \(2\), since \(2\) is a factor of \(4\))

Thus, \(f(x) = 25x^4 - 7x^2 + x + 4\) is the correct function.

Answer:

• (A) \(f(x) = 4x^4 - 7x^2 + x + 25\)
• (B) \(f(x) = 9x^4 - 7x^2 + x + 10\)
• (C) \(f(x) = 10x^4 - 7x^2 + x + 9\)
• (D) \(f(x) = 25x^4 - 7x^2 + x + 4\) (Correct answer)