Show that determinant is divisible by f(x)

Question

Let $\alpha$ be a repeated root of the quadratic equation $f(x)=0$ and $A(x),B(x),C(x) $be polynomials of degree 3,4 and 5 respectively.Then show that

\begin{vmatrix} A(x) & B(x) & C(x) \\ A(\alpha ) & B(\alpha ) & C(\alpha ) \\ A'(\alpha ) & B'(\alpha ) & C'(\alpha) \end{vmatrix}

is divisible by $f(x)$ where a dash denotes derivative w.r.t x.

Answer 1

Let the determinant be a function of $x$ :

$$g(x)=\begin{vmatrix} A(x) & B(x) & C(x) \\ A(\alpha ) & B(\alpha ) & C(\alpha ) \\ A'(\alpha ) & B'(\alpha ) & C'(\alpha) \end{vmatrix}$$

To show that $f(x)$ divides $g(x)$ we need to show that $\alpha$ is a double root for $g$ . This is equivalent with $g(\alpha)=g'(\alpha)=0$ .

But $$g(\alpha)=\begin{vmatrix} A(\alpha) & B(\alpha) & C(\alpha) \\ A(\alpha ) & B(\alpha ) & C(\alpha ) \\ A'(\alpha ) & B'(\alpha ) & C'(\alpha) \end{vmatrix}=0$$ because the determinant has the first two rows the same .

Also it's not hard to notice (just expand the determinant and then differentiate ) that :

$$g'(x)=\begin{vmatrix} A'(x) & B'(x) & C'(x) \\ A(\alpha ) & B(\alpha ) & C(\alpha ) \\ A'(\alpha ) & B'(\alpha ) & C'(\alpha) \end{vmatrix}$$ so using the same argument :

$$g'(\alpha)=\begin{vmatrix} A'(\alpha) & B'(\alpha) & C'(\alpha) \\ A(\alpha ) & B(\alpha ) & C(\alpha ) \\ A'(\alpha ) & B'(\alpha ) & C'(\alpha) \end{vmatrix}=0$$ because the first and last row are the same .

Answer 2

Hint as it has repeated root so it should be a perfect square so $f(x)=c(x\pm \alpha)^2$ now determinant to be divisible by it $\alpha$ should be the root of the epansion of determinant now as $\alpha $ is one of the root of $f(x)=0$ so we can say $x=\alpha$ thus determinant becomes $0$ after substituting $x=\alpha$ in $R_1$ thus determinant is divisible by $c(x\pm \alpha)^2$