Factoring a quadratic polynomial in standard form

\[\begin{align*} (qx+r)(sx+t)&=(qx)(sx)+(qx)(t)+(r)(sx)+(r)(t)\\ &=qsx^2+qtx+rsx+rt\\ &=(qs)x^2+(qt+sr)x+rt \end{align*}\] \[ax^2+bx+c=(qx+r)(sx+t)\]

if and only if

\[\begin{align*} a&=qs\\ b&=qt+sr\\ c&=rt \end{align*}\]

Let

\[m = qt\]

and

\[n = sr.\]

Then

\[mn=ac,\qquad m+n=b\]

and

\[\begin{align*} ax^2+bx+c&=ax^2+mx+nx+c\\ &=(qs)x^2+qtx+srx+rt\\ &=qx(sx+t)+r(sx+t)\\ &=(qx+r)(sx+t). \end{align*}\]

If \(mn=ac\) and \(m+n=b\) then \(n=b-m\) and so

\[m(b-m)=ac\]

i.e.

\[m^2-bm+ac=0\]

\[15x^2+61x+34\]