Question

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

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: Determine possible roots

Using the Rational Root Theorem, the possible rational roots \( \frac{p}{q} \) are given by \( \frac{\text{factor of }a_0}{\text{factor of }a_n} \). Since the factor of \( a_n = 1 \) is \( \pm1 \), the possible rational roots are \( \frac{\pm1}{\pm1},\frac{\pm2}{\pm1},\frac{\pm3}{\pm1},\frac{\pm6}{\pm1} \), which simplifies to \( \pm1,\pm2,\pm3,\pm6 \).

Answer:

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