Prove the given results 
Let $f(x)=a x^{2}+b x+c$ be a quadratic polynomial with integral coefficients, where $a \neq 0$. Show that
(i) if $f(x)$ is factorisable into linear factors with integral coefficients, then there are integers $d$ and $e$ such that
$$
d+e=b \text { and } d e=a c \text {; }
$$
and (ii) if integers $d$ and $e$ can be found such that (1) holds, then
$$
f(x)=\frac{(a x+d)}{g} \frac{(a x+e)}{a / g},
$$
where $g$ is the g.c.d. of $a$ and $d$ and each of the linear factors has integral coefficients.


After taking hints from the amswers and comments below, I have tried solving this question. Please let me know if it's correct or not. From eyeballfrog's idea, let's assume there are integers $p,q,r,s$ such that
$$
f(x) = ax^2 + b x + c = (px + q)(rx + s) = prx^2 + (ps+qr)x + qs.
$$. By comparing, we get, $a=pr, b=ps+qr, c=qs$. As $ps,qr$ are integers, we can let $ps=d, qr=e$, and from there we get $b=d+e$, and $de=ac$. Now for the second part, $\begin{aligned} f(x) &=a x^{2}+b x+c \\ &=\frac{a^{2} x^{2}+a b x+a c}{a} \\ &=\frac{a^{2} x^{2}+a b x+a c}{g \cdot \frac{a}{g}} \\ &=\frac{a^{2} x^{2}+a x(e+d)+a c}{g \cdot \frac{a}{g}} \\ &=\frac{a^{2} x^{2}+a x e+a x d+p q r s}{g \cdot \frac{a}{g}} \\ &=\frac{a^{2} x^{2}+a x e+a x d+d e}{g \cdot \frac{a}{g}} \end{aligned}$ $=\frac{a x(a x+e)+d(a x+e)}{g \cdot \frac{a}{g}}$
$=\frac{(a x+d)}{g} \cdot \frac{(a x+e)}{\frac{a}{g}}$.
 A: For part (i), we assume that $f$ has linear integer factors. That is, there are integers $p,q,r,s$ such that
$$
f(x) = ax^2 + b x + c = (px + q)(rx + s) = prx^2 + (ps+qr)x + qs.
$$
Can you find the desired integers $d$ and $e$ from here?
For part (ii), multiplying out the product shows it's equal to $f$. So we need to verify that all the coefficients are integers. Since $g =\gcd(a,d)$, $a/g$, $d/g$, and $a/(a/g) = g$ are clearly integers. What about $e/(a/g)$? Hint: you'll need the fact that $a/g$ and $d/g$ have no common factors.
A: For my proof, I first took the equations as assumptions and proved them to indeed be true.
$ f(x) = a (x-\alpha) (x-\beta) $, now.
$ d + e = b = -a (\alpha + \beta) $, $ de = ac = a^2 \alpha \beta $
Hence, if $ d = -a * \alpha $, $ e = -b * \beta $, they satisfy both equations. And since they add and multiply to become an integer hence they must be integers too.
In the second part of the question the g cancels out, and if you expand the brackets and put the values of $d+e$ and $de$ you will get back $f(x)$.
