Show that if $ax^2+bx+c$ can be factored such that the factors have integer coefficients, then there exists integers $u$ and $v$ such that $u+v=b$ and $uv=ac$.
This problem regards the "diamond method" which was not covered in class. It is used to factor $ax^2+bx+c$. First we find integers $u,v$ such that $u+v=b$ and $uv=ac$. Then we factor $ax^2+ux+vx+c$ by grouping.
2 Answers
Suppose that $ax^2+bx+c$ can be factored such that the factors have integer coefficients, so $ax^2+bx+c = (mx+n)(px+q)$. Then we have
$$mp=a\;,\qquad nq=c\;,\qquad np+mq = b\;.$$
Multiplying the first two equations together to get $mpnq=ac$, and rearranging we see that
$$(np)(mq) = ac\;, \qquad (np)+(mq) = b\;.$$
So $np = u$ and $mq=v$ are the desired integers.
So we have that $p(x) = ax^2 + bx + c$ can be factored degree one polynomial integer coefficients so:
$$p(x) = (kx + p)(mx + q)$$
But,
$$kxmx = kmx^2 = ax^2 \implies a = km$$
Furthermore, $$pq = c$$
So say that $u = kq$ and $v = mp$. Then:
$$kxq + mxp = (kq + mp)x = bx \implies (u+v) = b$$
