Question

Let $\alpha$ be a repeated root of the quadratic equation $f\left(x\right)=0$ and A(x), B(x), C(x) be polynomials of degree 3, 4 and 5 respectively, then show
$\Delta \left(x\right)=\left|\begin{array}{ccc}A\left(x\right)& B\left(x\right)& C\left(x\right)\\ A\left(\alpha \right)& B\left(\alpha \right)& C\left(\alpha \right)\\ A^{\prime }\left(\alpha \right)& B\left(\alpha \right)& C^{\prime }\left(\alpha \right)\end{array}\right|$ is divisible by f(x).

Answer 1

Let $\Delta \left(x\right)=\left|\begin{array}{ccc}A\left(x\right)& B\left(x\right)& C\left(x\right)\\ A\left(\alpha \right)& B\left(\alpha \right)& C\left(\alpha \right)\\ A^{\prime }\left(\alpha \right)& B\left(\alpha \right)& C^{\prime }\left(\alpha \right)\end{array}\right|$
$\therefore {\Delta }^{\prime }\left(x\right)=\left|\begin{array}{ccc}{A}^{\prime }\left(x\right)& B^{\prime }\left(x\right)& C^{\prime }\left(x\right)\\ A\left(\alpha \right)& B\left(\alpha \right)& C\left(\alpha \right)\\ A^{\prime }\left(\alpha \right)& B^{\prime }\left(\alpha \right)& C^{\prime }\left(\alpha \right)\end{array}\right|+0+0$
(Since $R_2$ and $R_3$ does not contains x, i.e., constant)
Now
$\Delta \left(\alpha \right)=\left|\begin{array}{ccc}A\left(\alpha \right)& B\left(\alpha \right)& C\left(\alpha \right)\\ A\left(\alpha \right)& B\left(\alpha \right)& C\left(\alpha \right)\\ A^{\prime }\left(\alpha \right)& B^{\prime }\left(\alpha \right)& C^{\prime }\left(\alpha \right)\end{array}\right|=0$
($\because R_1$ and $R_2$ are identical)
and ${\Delta }^{\prime }\left(\alpha \right)=\left|\begin{array}{ccc}{A}^{\prime }\left(\alpha \right)& B^{\prime }\left(\alpha \right)& C^{\prime }\left(\alpha \right)\\ A\left(\alpha \right)& B\left(\alpha \right)& C\left(\alpha \right)\\ A^{\prime }\left(\alpha \right)& B^{\prime }\left(\alpha \right)& C^{\prime }\left(\alpha \right)\end{array}\right|=0$ ($\because R_1$ and $R_3$
are identical )
It means that $\Delta$(x) is divisible by $(x-\alpha {)}^2$ which is repeated fractor of quadratic equation $f\left(x\right)=0$.
Hence $\Delta$(x) is divisible by $f\left(x\right)$.