Determinant composed from polynomials $p_1(x) = x + a$ and $p_2(x) = x^2 + bx + c$ Let $p_1(x) = x + a$ and $p_2(x) = x^2 + bx + c$ be two polynomials with real
coefficients, and $x_1$ and $x_2$ be two arbitrary real numbers. Consider the
following determinant $$D(x) = \begin{vmatrix} 
  1     & p_1(x_1) & p_2(x_1)\\ 
  1 & p_1(x_2) & p_2(x_2)\\
1 & p_1(x) & p_2(x)\
\end{vmatrix}
$$
Show that $D(x) = m(x-x_1)(x-x_2)$
 A: Hint: $D(x)$ is a $2^{nd}$ degree polynomial in $x\,$, and $D(x_1)=D(x_2)=0\,$.
A: To the determinant $\Delta$, we perform the row operations $R_2 \to R_1-R_2$ and $R_3 \to R_1-R_3$. Then we have $$\Delta = \begin {vmatrix}
1 & p_1 (x_1) & p_2 (x_1)\\
0 & p_1 (x_2)-p_1 (x_1) & p_2 (x_2)-p_2 (x_1)\\
0 & p_1 (x)-p_1 (x_1) & p_2 (x)-p_2 (x_1)\
\end {vmatrix}$$ Then expanding via $R_1$, we get, $$\Delta = (x_2-x_1)[x^2-x_1^2 +b (x-x_1)] -(x-x_1)[x_2^2-x_1^2 +b(x_2-x_1)] =(x-x_2)(x_2-x_1)(x-x_1) $$ Hope it helps. 
A: Note that since $\det$ is multi linear and alternating, we have
\begin{eqnarray}
D(x) &=& \det \begin{bmatrix} 
  1     & x_1+a & x_1^2+bx_1+c \\ 
  1 & x_2+a & x_2^2+bx_2+c \\
1 & x+a & x^2+bx+c \
\end{bmatrix} \\
&=& \det \begin{bmatrix} 
  1     & x_1 & x_1^2+bx_1+c \\ 
  1 & x_2 & x_2^2+bx_2+c \\
1 & x & x^2+bx+c \
\end{bmatrix} \\
&=& \det \begin{bmatrix} 
  1     & x_1 & x_1^2+bx_1 \\ 
  1 & x_2 & x_2^2+bx_2 \\
1 & x & x^2+bx \
\end{bmatrix} \\
&=& \det \begin{bmatrix} 
  1     & x_1 & x_1^2 \\ 
  1 & x_2 & x_2^2 \\
1 & x & x^2 \
\end{bmatrix} \\
\end{eqnarray}
We see from this that $D$ is a second degree polynomial in $x$ and
$D(x_1) = D(x_2) = 0$. We can read off the coefficient $m$ as the coefficient of $x^2$ above, $m=x_2-x_1$.
(As Joonas notes below, the above is a Vandermonde matrix which has a well
known formula for the determinant.)
