Question

roots of a trinomial
\[ax^3+bx^2+cx+d=0\]

let the roots be \(p,q,r\)

When \(x\) is equal a root the equation equals zero

\[(x-p)(x-q)(x-r)=0\]
\[(x-p)(x^2-(q+r)x+qr)=0\]
\[x\left(x^2-(q+r)x+qr\right)-p\left(x^2-(q+r)x+qr\right)=0\]
\[\left(x^3-(q+r)x^2+(qr)x\right)-\left(px^2-p(q+r)x-qpr\right)=0\]
\[x^3-(p+q+r)x^2+(qr+pq+pr)x-pqr=0\]
comparing with
\[x^3+\frac bax^2+\frac cax+\frac da=0\]

\[-(p+q+r)=\frac ba\qquad\text{the sum of the roots is } -\frac ba\]
\[(qr+pq+pr)=\frac ca\qquad\text{sum of products of pairs of the roots is } \frac ca\]
\[-pqr=\frac da\qquad \text{the product of the roots is} -\frac da \]