"Find all integers $n$ such that the quadratic $7x^2 + nx - 11$ can be expressed as the product of two linear factors with integer coefficients.
In a quadratic in the form of $ax^2+bx+c$, the product of the roots of the quadratic equal the constant ($c$). Since there can only be integer coefficients, the only factors of $c$ ($11$) are $-1$ and $11$, and $1$ and $-11$. So we can put these in.
$(7x+1)(x-11)=$
$7x^2+x-77x-11=$
$7x^2-76x-11$
$N$ can be $-76$.
$(7x-1)(x+11)=$
$7x^2-x+77x-11=$
$7x^2+76x-11$
$N$ can be $76$. There is also another change you can make with the coefficient of $x^2$. You can move it over to the other side, so now we have....
$(x+1)(7x-11)=$
$7x^2+7x-11x-11=$
$7x^2-4x-11$
$N$ can be $-4$.
$(x-1)(7x+11)=$
$7x^2-7x+11x-11=$
$7x^2+4x-11.$
$N$ can be $4$.
So now we have these values of $n$: $-76$, $76$, $-4$, $4$.
Basically I am looking for a critique of my solution if it is correct.
I think it is correct, but it would be easier to use that the discriminant must be a square number, i.e. $n^2+308$ must be a square number. This has the solutions $n=\pm 4, \pm 76$.