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: Build the quadratic factor from the given complex root.
Since the coefficients are real, \(3-i\sqrt{6}\) is also a root, giving the factor \((x-3)^2+6=x^2-6x+15\).

Step 2: Divide it out of the quartic.
Dividing \(4x^4-24x^3+57x^2+18x-45\) by \(x^2-6x+15\): the first quotient term is \(4x^2\), and subtracting \(4x^2(x^2-6x+15)=4x^4-24x^3+60x^2\) leaves \(-3x^2+18x-45\).
The next quotient term is \(-3\), and subtracting \(-3(x^2-6x+15)=-3x^2+18x-45\) leaves remainder \(0\).
So the quotient is \(4x^2-3\), meaning \(4x^4-24x^3+57x^2+18x-45=(x^2-6x+15)(4x^2-3)\).

Step 3: Find the product of the real roots.
The real roots come from \(4x^2-3=0\); by the ratio of constant to leading coefficient, their product is \(\frac{-3}{4}\).

Step 4: Conclusion.
\[ \boxed{-\frac{3}{4}} \]