The remainder when \(x^{3} – 2x + m\) is divided by \(x – 1\) is equal to the remainder when \(2x^{3} + x – m\) is divided by \(2x + 1\). Find the value of m.
- \(\frac{-7}{8}\)
- \(\frac{-3}{8}\)
-
\(\frac{1}{8}\)
- \(\frac{5}{8}\)
The correct answer is: C
Explanation
The remainder theorem states that if f(x) is divided by (x - a), the remainder is f(a).
\(f(x) = x^{3} - 2x + m\) divided by (x - 1), so that a = 1.
Remainder = \(f(1) = 1^3 - 2(1) + m = -1 + m\)
\(f(x) = 2x^{3} + x - m\) divided by (2x + 1), so that a = \(\frac{-1}{2}\)
\(f(\frac{-1}{2}) = 2(\frac{-1}{2}^{3}) + (\frac{-1}{2}) - m = \frac{-3}{4} - m\)
\(\implies m - 1 = \frac{-3}{4} - m\), collecting like terms,
\(2m = \frac{1}{4} \therefore m = \frac{1}{8}\)