Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable 
Prove that if $b^2-4ac=k^2$ (for some positive integer $k$ ) then $ax^2+bx+c$ is factorizable

I related this to the roots of equation: $ax^2+bx+c=0$ and using roots $x_1,x_2$:
$(x-x_1)(x-x_2)=0\implies(2ax+k-b)(2ax+k+b)=0\implies 4a^2x^2+4akx+k^2-b^2=0$ but this seems has nothing to do with original equation.Any better way??!!  
Under what conditions can we be sure the factorization involves just integers?? 
 A: $$ax^2+bx+c=a\left(\left(x-\frac{b}{a}\right)^2-\frac{b^2-4ac}{a^2}\right)=a\left(\left(x-\frac{b}{a}\right)^2-\frac{k^2}{a^2}\right)$$
which is of the form $$\alpha ^2-\beta^2=(\alpha -\beta )(\alpha +\beta ).$$
A: First note that there is no mention of "$=0$" here. There is no equation, only the polynomial itself. That means you cannot multiply away the denominators (there is no $0$ on the right-hand side to absorb it). Of course, you have to "put it into an equation" to find $x_1$ and $x_2$, but for the factorisation itself, don't equationize it (I don't think that's a word...).
Also, the solution formula to said equation is $\frac{-b\pm k}{2a}$, not $\frac{-k\pm b}{2a}$, which means that your last expression should've been $4a^2x^2 + 4abx + b^2 - k^2$. If you swap back $k^2 = b^2 - 4ac$ and divide the whole thing by $4a$ again (you multiplied by $4a$ to take away the denominators), you might find that it has more to do with the original expression than you thought. Here is a full working-out, for completeness:
$$
a(x-x_1)(x-x_2) = a\left(x+\frac{b+k}{2a}\right)\left(x+\frac{b-k}{2a}\right)\\
= a\left(x + \frac{b}{2a}\right)^2 - a\left(\frac{k}{2a}\right)^2 = ax^2 + bx + \frac{b^2}{4a} - \frac{b^2 - 4ac}{4a} = ax^2 + bx + c
$$
A: Hint: using the quadratic formula to solve the equation
$$ax^2+bx+c=0$$ we get:
$$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ What can you now say about the solutions if you know that $b^2-4ac=k^2,k\in\mathbb N$?
