Question

If $x^3 - 6x^2 + 11x - 6$ is a prime number then the number of possible integral values of x is

Answer 1

0

Answer 2

Let $P(x) = x^3 - 6x^2 + 11x - 6$. We can factorize $P(x)$ as follows:

$P(x) = (x-1)(x-2)(x-3)$

We are given that $P(x)$ is a prime number. Let $P(x) = p$, where $p$ is a prime number. $p = (x-1)(x-2)(x-3)$. Since $x$ is an integer, the factors $(x-1), (x-2), (x-3)$ are integers.

We need to find the number of integral values of $x$ for which $P(x)$ is a prime number.

Consider the case where $x=1, 2, 3$. If $x=1$, $P(1) = (1-1)(1-2)(1-3) = 0$. If $x=2$, $P(2) = (2-1)(2-2)(2-3) = 0$. If $x=3$, $P(3) = (3-1)(3-2)(3-3) = 0$. Since 0 is not a prime number, $x=1, 2, 3$ are not solutions.

If $x > 3$, then $x-1, x-2, x-3$ are all positive integers greater than 0. Since $x-1 > x-2 > x-3$, we have $x-3 \ge 1$, $x-2 \ge 2$, $x-1 \ge 3$. The product of three integers greater than 1 cannot be a prime number.

If $x < 1$, then $x-1, x-2, x-3$ are all negative integers. Then $(x-1)(x-2)(x-3)$ is a negative number, which cannot be a prime number.

Thus, there are no integral values of $x$ for which $P(x)$ is a prime number. Therefore, the number of possible integral values of $x$ is 0.

Final Answer: The final answer is $\boxed{0}$