Let \(x\), \(y\), and \(z\) be the roots of \(f(t)=at^3+bt^2+ct+d\). We can divide by \(a\) on both sides here because it doesn't affect the roots. This gives us
\(\frac{f(t)}{a}=t^3+\frac{b}{a}t^2+\frac{c}{a}t+\frac{d}{a}.\)
By Vieta's Formulas, \(x+y+z=-\frac{b}{a}\), \(xy + xz + yz=\frac{c}{a}\), and \(xyz=-\frac{d}{a}\). Since \(x + y + z\), \(xy + xz + yz\), and \(xyz\) are all positive, \(\frac{b}{a}\) must be negative, \(\frac{c}{a}\) must be positive, and \(\frac{d}{a}\) must be negative.
For \(\frac{f(t)}{a}=0\), \(t\) has to be positive. This is because if \(t\) was negative, \(t^3\) would be negative, \(\frac{b}{a}t^2\) would be negative (\(t^2\) is positive but \(\frac{b}{a}\) is negative), \(\frac{c}{a}t\) would be negative, and \(\frac{d}{a}\) would be negative so \(\frac{f(t)}{a}\) would have to be negative. If \(t\) was \(0\), \(t^3\) would be \(0\), \(\frac{b}{a}t^2\) would be \(0\), \(\frac{c}{a}t\) would be \(0\), and \(\frac{d}{a}\) would be negative so \(\frac{f(t)}{a}\) would still be negative. Therefore, the only way for \(\frac{f(t)}{a}=0\) is for \(t\) to be positve. This means that all of \(f(t)\)'s roots (which are the same as \(\frac{f(t)}{a}\)'s roots) are positive so \(x, y,\) and \(z\) are all positive.
