Question

according to the rational root theorem, $-\frac{2}{5}$ is a potential rational root of which function?
$\bigcirc$ $f(x) = 4x^4 - 7x^2 + x + 25$
$\bigcirc$ $f(x) = 9x^4 - 7x^2 + x + 10$
$\bigcirc$ $f(x) = 10x^4 - 7x^2 + x + 9$
$\bigcirc$ $f(x) = 25x^4 - 7x^2 + x + 4$

Explanation:

Step1: Recall Rational Root Theorem

The Rational Root Theorem states that any potential rational root \(\frac{p}{q}\) of a polynomial \(a_nx^n + \dots + a_1x + a_0\) has \(p\) as a factor of the constant term \(a_0\) and \(q\) as a factor of the leading coefficient \(a_n\). Here, \(\frac{p}{q}=-\frac{2}{5}\), so \(p = \pm2\) (factor of constant term) and \(q=\pm5\) (factor of leading coefficient).

Step2: Analyze each function

• First function (\(f(x) = 4x^4 - 7x^2 + x + 25\)): Leading coefficient \(a_n = 4\), constant term \(a_0 = 25\). Factors of \(25\): \(\pm1,\pm5,\pm25\); factors of \(4\): \(\pm1,\pm2,\pm4\). \(q = 5\) is not a factor of \(4\), so \(-\frac{2}{5}\) not potential root.
• Second function (\(f(x) = 9x^4 - 7x^2 + x + 10\)): Leading coefficient \(a_n = 9\), constant term \(a_0 = 10\). Factors of \(10\): \(\pm1,\pm2,\pm5,\pm10\); factors of \(9\): \(\pm1,\pm3,\pm9\). \(q = 5\) is not a factor of \(9\), so \(-\frac{2}{5}\) not potential root.
• Third function (\(f(x) = 10x^4 - 7x^2 + x + 9\)): Leading coefficient \(a_n = 10\), constant term \(a_0 = 9\). Factors of \(9\): \(\pm1,\pm3,\pm9\); factors of \(10\): \(\pm1,\pm2,\pm5,\pm10\). \(q = 5\) is a factor of \(10\), \(p = 2\) is a factor of \(9\)? No, \(2\) is not a factor of \(9\). Wait, no: \(p\) is factor of constant term (\(9\))? Wait, no: \(p\) is factor of \(a_0\) (constant term), \(q\) is factor of \(a_n\) (leading coefficient). So \(p\) should be factor of \(a_0\), \(q\) factor of \(a_n\). Here \(a_0 = 9\), factors of \(9\) are \(\pm1,\pm3,\pm9\); \(2\) is not a factor of \(9\), so \(p = 2\) is invalid. Wait, mistake: in \(-\frac{2}{5}\), \(p = - 2\) (so factor of \(a_0\)), \(q = 5\) (factor of \(a_n\)). So for \(f(x)=10x^4 - 7x^2 + x + 9\), \(a_0 = 9\), \(2\) is not a factor of \(9\), so no.
• Fourth function (\(f(x) = 25x^4 - 7x^2 + x + 4\)): Leading coefficient \(a_n = 25\), constant term \(a_0 = 4\). Factors of \(4\): \(\pm1,\pm2,\pm4\) (so \(p = - 2\) is a factor); factors of \(25\): \(\pm1,\pm5,\pm25\) (so \(q = 5\) is a factor). Thus, \(-\frac{2}{5}\) is a potential rational root since \(p\) (factor of \(a_0\)) and \(q\) (factor of \(a_n\)).

Answer:

D. \(f(x) = 25x^4 - 7x^2 + x + 4\) (assuming the options are labeled A, B, C, D as first, second, third, fourth; so the fourth option: \(f(x) = 25x^4 - 7x^2 + x + 4\))