We are asked to find the product of the real roots of the quartic polynomial $x^4 + 3x^3 + 5x^2 + 21x - 14 = 0$.

1. Problem Description

We are asked to find the product of the real roots of the quartic polynomial $x^4 + 3x^3 + 5x^2 + 21x - 14 = 0$.

2. Solution Steps

We can try to factor the quartic polynomial. Let's attempt to factor it into two quadratic polynomials.

$x^4 + 3x^3 + 5x^2 + 21x - 14 = (x^2 + ax + b)(x^2 + cx + d) = x^4 + (a+c)x^3 + (ac+b+d)x^2 + (ad+bc)x + bd$

Comparing coefficients, we have:

$a+c = 3$

$ac+b+d = 5$

$ad+bc = 21$

$bd = -14$

Since $bd = -14$, let's try $b = -2$ and $d = 7$. Then we have:

$a+c = 3$

$ac - 2 + 7 = 5 \Rightarrow ac = 0$

$7a - 2c = 21$

Since $ac=0$, either $a=0$ or $c=0$.

If $a=0$, then $c=3$, and $-2c = 21$, so $-2(3) = 21$, which is $-6 = 21$, which is not true.

If $c=0$, then $a=3$, and $7a = 21$, so $7(3) = 21$, which is true.

Therefore, we have $a=3$, $b=-2$, $c=0$, $d=7$.

So the factorization is:

$x^4 + 3x^3 + 5x^2 + 21x - 14 = (x^2 + 3x - 2)(x^2 + 7) = 0$

We need to find the real roots.

$x^2 + 7 = 0$ has no real roots since $x^2 = -7$, which means $x = \pm i\sqrt{7}$.

$x^2 + 3x - 2 = 0$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-2)}}{2(1)}$

$x = \frac{-3 \pm \sqrt{9 + 8}}{2}$

$x = \frac{-3 \pm \sqrt{17}}{2}$

So the real roots are $x_1 = \frac{-3 + \sqrt{17}}{2}$ and $x_2 = \frac{-3 - \sqrt{17}}{2}$.

The product of the real roots is $x_1 x_2 = \frac{(-3 + \sqrt{17})(-3 - \sqrt{17})}{4} = \frac{(-3)^2 - (\sqrt{17})^2}{4} = \frac{9 - 17}{4} = \frac{-8}{4} = -2$.

3. Final Answer

-2