Question

If 2, 3, 6 are the roots of the polynomial \(f(x) = x^3+ax^2+bx+c\), where \(a,b,c \in C\). Then the value of a--c is

• A. -11
• B. 36
• C. 25
• D. 11

Hint:

For a cubic polynomial \(x^3+ax^2+bx+c=0\) with roots \(\alpha, \beta, \gamma\):
\(\alpha+\beta+\gamma = -a\)
\(\alpha\beta+\beta\gamma+\gamma\alpha = b\)
\(\alpha\beta\gamma = -c\)
Interpret "a--c" carefully; it most likely means \(a-c\).

Answer 1

The Correct Option is C

Given the roots of the polynomial \(f(x) = x^3+ax^2+bx+c\) are 2, 3, and 6. From Vieta's formulas for a cubic polynomial \(x^3 - (\text{sum of roots})x^2 + (\text{sum of products of roots taken two at a time})x - (\text{product of roots}) = 0\): The polynomial can also be written as \((x-2)(x-3)(x-6) = 0\). Comparing \(x^3+ax^2+bx+c\) with standard \(x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta+\beta\gamma+\gamma\alpha)x - \alpha\beta\gamma\): Sum of roots: \(2+3+6 = 11\). So, \(-a = 11 \Rightarrow a = -11\). Sum of products of roots taken two at a time: \((2)(3) + (3)(6) + (6)(2) = 6 + 18 + 12 = 36\). So, \(b = 36\). Product of roots: \((2)(3)(6) = 36\). So, \(-c = 36 \Rightarrow c = -36\). We need to find the value of "a--c". If "a--c" means \(a-c\): \(a-c = (-11) - (-36) = -11 + 36 = 25\). This matches option (c) if "a--c" is \(a-c\). If "a--c" means \(a-(-c) = a+c\): \(a+c = (-11) + (-36) = -11 - 36 = -47\). This is not among the options. So, it is highly likely "a--c" is a typo for "a-c". \[ \boxed{25} \]