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# Number of positive integers x for which $f\left( x \right) = {x^3} - 8{x^2} + 20x - 13$ , is a prime number is:(A) $1$ (B) $2$ (C) $3$ (D) $4$ Verified
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Hint: In the given question, we are given a function in the variable x and we have to find the number of positive integers x for which the function assumes a prime number value. So, we first factorize the given function into its factors. We use the factor theorem to find the factors of the cubic polynomial function. Then, we analyze the product of the prime factors in order to make the value of the function a prime number. A thorough understanding of functions, division algorithms and its applications will be of great significance.

In the given problem, we are required to find the number of positive integers for which the function $f\left( x \right) = {x^3} - 8{x^2} + 20x - 13$ assumes a prime number value.
So, we first find the factors of the polynomial function $f\left( x \right) = {x^3} - 8{x^2} + 20x - 13$ using the factor theorem.
According to the factor theorem, if the value of the variable obtained on equating the divisor to zero is a root of the polynomial, then the divisor polynomial is a factor of the dividend polynomial.
Now, let us substitute the value of the variable x as $1$ in the function $f\left( x \right) = {x^3} - 8{x^2} + 20x - 13$ .
So, we get,
$\Rightarrow f\left( 1 \right) = {1^3} - 8{\left( 1 \right)^2} + 20\left( 1 \right) - 13$
Simplifying the expression further, we get,
$\Rightarrow f\left( 1 \right) = 1 - 8 + 20 - 13$
$\Rightarrow f\left( 1 \right) = 0$
Since the value of polynomial function $f\left( x \right) = {x^3} - 8{x^2} + 20x - 13$ for $x = 1$ is zero, then according to the factor theorem, $\left( {x - 1} \right)$ is a factor of the polynomial function $f\left( x \right) = {x^3} - 8{x^2} + 20x - 13$ .
Now, factoring the polynomial function to factor out $\left( {x - 1} \right)$ , we get,
$\Rightarrow f\left( x \right) = {x^3} - {x^2} - 7{x^2} + 7x + 13x - 13$
$\Rightarrow f\left( x \right) = {x^2}\left( {x - 1} \right) - 7x\left( {x - 1} \right) + 13\left( {x - 1} \right)$
Taking $\left( {x - 1} \right)$ common from al the terms, we get,
$\Rightarrow f\left( x \right) = \left( {{x^2} - 7x + 13} \right)\left( {x - 1} \right)$
Now, for the function $f\left( x \right) = {x^3} - 8{x^2} + 20x - 13$ to be a prime number, one of the factors of the function must be $1$ and the other factor must be a prime number.
So, there are two cases.
Case $1$ : When $\left( {x - 1} \right) = 1$ and $\left( {{x^2} - 7x + 13} \right)$ is a prime factor.
So, $\left( {x - 1} \right) = 1$
$\Rightarrow x = 2$
Hence, the value of x comes out to be $2$ .
Now, we find the value of $\left( {{x^2} - 7x + 13} \right)$ for $x = 2$ .
$\Rightarrow {2^2} - 7\left( 2 \right) + 13$
$\Rightarrow 4 - 14 + 13 = 3$
Therefore, $x = 2$ is a positive integral value of x for which the function $f\left( x \right) = {x^3} - 8{x^2} + 20x - 13$ is a prime number.
Case $2$ : When $\left( {{x^2} - 7x + 13} \right) = 1$ and $\left( {x - 1} \right)$ is a prime factor.
So, $\left( {{x^2} - 7x + 13} \right) = 1$
$\Rightarrow {x^2} - 7x + 12 = 0$
Solving the quadratic equation using splitting the middle term method, we get,
$\Rightarrow {x^2} - 3x - 4x + 12 = 0$
$\Rightarrow x\left( {x - 3} \right) - 4\left( {x - 3} \right) = 0$
Taking $\left( {x - 3} \right)$ common from the terms, we get,
$\Rightarrow \left( {x - 4} \right)\left( {x - 3} \right) = 0$
So, $x = 3$ or $x = 4$
Now, we find the value of $\left( {x - 1} \right)$ for $x = 3$ and $x = 4$ .
So, $\left( {x - 1} \right) = 2$ for $x = 3$ .
Also, $\left( {x - 1} \right) = 3$ for $x = 4$ .
Therefore, $x = 3$ and $x = 4$ are the positive integral values of x for which the function $f\left( x \right) = {x^3} - 8{x^2} + 20x - 13$ is a prime number.
Hence, the total number of positive integers for which $f\left( x \right) = {x^3} - 8{x^2} + 20x - 13$ is a prime number is $3$ .
Therefore, option (C) is correct.
So, the correct answer is “Option C”.

Note: One must know the significance of the factor theorem in order to solve such problems. Substitution of a variable involves putting a certain value in place of the variable. That specified value may be a certain number or even any other variable. We must take care of the calculations while doing such problems. We must know the basics of the number system to solve such problems. Prime numbers are basic building blocks of arithmetic.