Question

The product of real roots of the equation \[ 4x^4 - 24x^3 + 57x^2 + 18x - 45 = 0 \] if one of the roots is \(3 + i\sqrt{6}\) is:

• A. \(-5/16\)
• B. \(5/16\)
• C. \(3/4\)
• D. \(-3/4\)

Hint:

For polynomials with complex roots, factor out quadratic of conjugate pair to find remaining real roots; use product formula \(c/a\) for product of roots.

Answer 1

The Correct Option is D

Step 1: Use complex conjugate root property.
Since coefficients are real, if \(3 + i\sqrt{6}\) is a root, its conjugate \(3 - i\sqrt{6}\) is also a root.

Step 2: Factor out quadratic corresponding to complex roots.
\[ (x - (3+i\sqrt{6}))(x - (3-i\sqrt{6})) = x^2 - 6x + 15 \]

Step 3: Perform polynomial division.
Divide \(4x^4 - 24x^3 + 57x^2 + 18x - 45\) by \(x^2 - 6x + 15\) to get remaining quadratic \(4x^2 + 3\).

Step 4: Find product of real roots.
The remaining quadratic: \(4x^2 + 3 = 0\) yields real roots \(x_1, x_2\) (check signs). Product of roots = \(\frac{c}{a} = \frac{-3}{4} = -3/4\)

Step 5: Final conclusion.
\[ \boxed{-3/4} \]