Question

list all possible roots of the polynomial $x^{3}-6x^{2}+11x - 6$ \\(\pm1,\pm6\\) \\(\pm1,\pm2,\pm3,\pm6\\) \\(\pm\frac{1}{6},\pm\frac{1}{3},\pm\frac{1}{2}\\) \\(\pm2,\pm3\\)

Explanation:

Step1: Recall Rational Root Theorem

The Rational Root Theorem states that any possible rational root, \( \frac{p}{q} \), of a polynomial equation \( a_nx^n + a_{n - 1}x^{n - 1}+\dots+a_1x + a_0 = 0 \) (where \( a_n
eq0 \)) is a factor of the constant term \( a_0 \) divided by a factor of the leading coefficient \( a_n \).

For the polynomial \( x^3 - 6x^2 + 11x - 6 \), the leading coefficient \( a_n=1 \) (the coefficient of \( x^3 \)) and the constant term \( a_0 = - 6 \).

Step2: Find factors of \( a_0 \) and \( a_n \)

• Factors of the constant term \( a_0=-6 \) are \( \pm1,\pm2,\pm3,\pm6 \).
• Factors of the leading coefficient \( a_n = 1 \) are \( \pm1 \).

Step3: Apply the Rational Root Theorem

Using the formula \( \frac{p}{q} \), where \( p \) is a factor of \( a_0 \) and \( q \) is a factor of \( a_n \), we have \( \frac{p}{q}=\frac{\pm1,\pm2,\pm3,\pm6}{\pm1}=\pm1,\pm2,\pm3,\pm6 \).

Answer:

\( \pm1,\pm2,\pm3,\pm6 \) (corresponding to the option: \( \boldsymbol{\pm1,\pm2,\pm3,\pm6} \))