Which of the followings gives the product of the real roots of the equation $x^4+3x^3+5x^2 + 21x -14=0$?

$ 
\textbf{(A)}\ -2
\qquad\textbf{(B)}\ 7
\qquad\textbf{(C)}\ -14 
\qquad\textbf{(D)}\ 21
\qquad\textbf{(E)}\ \text{None of the preceding}
$
1. Given the polynomial equation:
   \[
   x^4 + 3x^3 + 5x^2 + 21x - 14 = 0
   \]
   We need to find the product of the real roots of this equation.

2. We start by attempting to factorize the polynomial. Notice that we can group terms to facilitate factorization:
   \[
   x^4 + 3x^3 + 5x^2 + 21x - 14 = (x^4 + 5x^2 - 14) + (3x^3 + 21x)
   \]

3. Factor out common terms from each group:
   \[
   (x^2 + 7)(x^2 - 2) + 3x(x^2 + 7)
   \]

4. Notice that both groups contain the factor \((x^2 + 7)\):
   \[
   (x^2 + 7)(x^2 - 2) + 3x(x^2 + 7) = (x^2 + 7)(x^2 - 2 + 3x)
   \]

5. Simplify the expression inside the parentheses:
   \[
   (x^2 + 7)(x^2 + 3x - 2)
   \]

6. Now, we need to find the real roots of the quadratic equation \(x^2 + 3x - 2 = 0\). We use the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \(a = 1\), \(b = 3\), and \(c = -2\).

7. Calculate the discriminant:
   \[
   \Delta = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot (-2) = 9 + 8 = 17
   \]

8. Since the discriminant is positive, the quadratic equation has two distinct real roots:
   \[
   x = \frac{-3 \pm \sqrt{17}}{2}
   \]

9. The product of the roots of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(\frac{c}{a}\). For the equation \(x^2 + 3x - 2 = 0\), the product of the roots is:
   \[
   \frac{c}{a} = \frac{-2}{1} = -2
   \]

10. Therefore, the product of the real roots of the original polynomial equation is \(\boxed{-2}\).