WAEC - Further Mathematics (2016 - No. 9)

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.
\(\frac{-7}{8}\)
\(\frac{-3}{8}\)
\(\frac{1}{8}\)
\(\frac{5}{8}\)

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,

\(2m = \frac{1}{4} \therefore m = \frac{1}{8}\)

Comments (0)

Advertisement