$x^n +a_{n−1}x^{n−1} + · · · + a_1x + a_0 ∈ Z[x]$, and it has a linear factor which must be of the form $x − a$ with $a$ dividing $a_0$. Prove that if $x^n +a_{n−1}x^{n−1} + · · · + a_1x + a_0 ∈ Z[x]$, and it has a
linear factor, then this linear factor must be of the form $x − a$ with
$a$ dividing $a_0$.
I'm really struggling to understand this question, any help would be great.
If f(x) = $x^n +a_{n−1}x^{n−1} + · · · + a_1x + a_0 $ has a linear factor, then the following equation is true :
\begin{align} x^n +a_{n−1}x^{n−1} + · · · + a_1x + a_0   = (bx-a)(g(x)+c) =b(\frac{x}{b}-\frac{a}{b})(g(x)+c)\end{align}
where b$\neq0$ and g(x) is a n-1 order function that does not have a constant term.
For this equation, RHS constant term is  $\frac{ac}{b}$, the linear term is $\frac{x}{b}$, only when b=1 this factor becomes  (x-a), in which case   a$_0$ $\to$ ac=a$_0$, or  c=$\frac{a_0}{a}$ . So, the statement is not true.
If f(x) = $x^n +a_{n−1}x^{n−1} + · · · + a_1x + a_0 $ has a linear factor, then the following equation is true :
\begin{align} x^n +a_{n−1}x^{n−1} + · · · + a_1x + a_0   = (bx-a)(g(x)+c) =b(\frac{x}{b}-\frac{a}{b})(g(x)+c)\end{align}
where b$\neq0$ and g(x) is a n-1 order function that does not have a constant term.
For this equation, RHS constant term is  $\frac{ac}{b}$, the linear term is $\frac{x}{b}$, only when b=1 this factor becomes  (x-a), in which case   a$_0$ $\to$ ac=a$_0$, or  c=$\frac{a_0}{a}$ . So, the statement is not true.