Question

X^3 - 8x^3 +20x -13 is a prime number.(x is natural number).find x​

Answer 1

$\Hugex=2,3,4$

$\Large\textrm{Given: -}$

"$x^{3}-8x^{2}+20x-13$ is a prime. Find $x$ where $x$ is natural."

Let us denote the polynomial $f(x)$.

 

$\textrm{Let us substitute}$

$(x=1)$

$f(1)=1-8+20-13=0$



$\Large\boxed{\textrm{Factor Theorem}}$

"If $f(a)=0$, the polynomial $f(x)$ is divisible by $(x-a)$."

$\Large\textrm{We know that: -}$

$f(1)=0$

$\textrm{by factor theorem}$

$f(x)=(x-1)Q(x)$

$\textrm{applying division}$

$Q(x)=x^{2}-7x+13$

Since the prime $f(x)$ consists of two factors, one factor should attain 1.

$\large\textrm{It follows that: -}$

$(x>0)$

[tex]\text{$\begin{cases} & x-1=1 \\ & x^{2}-7x+13>1 \end{cases}$ or $\begin{cases} & x-1>1 \\ & x^{2}-7x+13=1 \end{cases}$}[/tex]

$\text{ \begin{cases} & x=2 \\ & x<3,x>4 \end{cases} or \begin{cases} & x>2 \\ & x=3,4 \end{cases} }$

$\textrm{hence}$

$x=2,3,4$

But we don't know if the remaining factor is a prime.

$\textrm{So, verifying}$

$x=2 \Longrightarrow f(2)=3$

$x=3\Longrightarrow f(3)=2$

$x=4\Longrightarrow f(4)=3$

Hence, all numbers are prime.