Ytukyg

My Attempt at this is as follows:

Let $f(x) = a_0+a_1x+a_2x^2+...+a_mx^m$ $forall$ $a_i$ $in$ $F[X]$.

Given $f(a)=0$

$Rightarrow$ $$f(x) = (x-a)t(x) label{a}tag{1}$$ such that $deg(t(x))geq1$,

and since $deg(x-alpha)=1$, we can express $f(x)$ as a product of two functions such that their degree is not zero, hence $f(x)$ is reducible.

So for proving the above result is this approach allowed? Or should I use some other way or do I have to prove the equation which I used in $ref{a}$?

You should use polynomial division, otherwise you're simply stating what you want to prove. The division algorithm states that if $f(x)$ and $g(x)$ are polynomials, with $gne0$, then there exist unique polynomials $t(x)$ and $r(x)$ such that

1. 
   $f(x)=g(x)t(x)+r(x)$;


2. 
   $deg r(x)<deg g(x)$ (or $r=0$).

In this case we can take $g(x)=x-alpha$, so
$$
f(x)=(x-alpha)t(x)+r(x)
$$
where $r=0$ or has degree less than $deg(x-alpha)=1$. Therefore $r(x)=c$ is constant.

Evaluate at $alpha$:
$$
0=f(alpha)=(alpha-alpha)t(alpha)+c=c
$$
Hence $f(x)=(x-alpha)t(x)$. Since $deg f(x)>1$, $deg t(x)>0$, so $f$ is reducible.

Your proof is correct. I would not write “s.t. $deg t(x)geqslant1$”. It's more specific than that (although what you wrote is correct): $deg t(x)=m-1$. And $f(x)$ doesn't became reducible; it is reducible.

And the equality $(1)$ doesn't have to be proved (as part of this proof), since it is a standard theorem about polynomials.