To multiply the polynomials \( (3x - 8)(2x^2 + 4x - 9) \), we will distribute each term in the first polynomial to each term in the second polynomial.
Starting with \( 3x \):
1. \( 3x \cdot 2x^2 = 6x^3 \)
2. \( 3x \cdot 4x = 12x^2 \)
3. \( 3x \cdot -9 = -27x \)
Next, we multiply \( -8 \):
1. \( -8 \cdot 2x^2 = -16x^2 \)
2. \( -8 \cdot 4x = -32x \)
3. \( -8 \cdot -9 = 72 \)
Now, we combine all the products:
\[
6x^3 + 12x^2 - 27x - 16x^2 - 32x + 72
\]
Next, we combine like terms:
1. For \( x^3 \): \( 6x^3 \)
2. For \( x^2 \): \( 12x^2 - 16x^2 = -4x^2 \)
3. For \( x \): \( -27x - 32x = -59x \)
4. Constant term: \( 72 \)
Putting it all together, we rewrite the polynomial in descending order:
\[
6x^3 - 4x^2 - 59x + 72
\]
The coefficient of the \( x^3 \) term is \( 6 \).
Thus, the final answer for the coefficient of the \( x^3 \) term is:
\[
\boxed{6}
\]