WAEC Further Mathematics 2015 Paper

Concept:By the factor theorem, if $(ax + b)$ is a factor of a polynomial $f(x)$, then $f\left(-\frac{b}{a}\right) = 0$.Explanation:Let $f(x) = 6x^{4} + 2x^{3} + 15x + 5$.Test each option by substituting the root of each possible factor into $f(x)$.For option A, $(3x + 1) = 0$ gives $x = -\frac{1}{3}$.Then $f\left(-\frac{1}{3}\right) = 6\left(-\frac{1}{3}\right)^{4} + 2\left(-\frac{1}{3}\right)^{3} + 15\left(-\frac{1}{3}\right) + 5$.This equals $\frac{2}{27} - \frac{2}{27} - 5 + 5 = 0$.Since the result is zero, $(3x + 1)$ divides the polynomial without a remainder.Checking the remaining options: $f(-1) = 6 - 2 - 15 + 5 = -6 \neq 0$.$f\left(-\frac{1}{2}\right) \neq 0$ and $f(-2) = 96 - 16 - 30 + 5 = 55 \neq 0$.Only option A satisfies the factor theorem.Answer:A. $3x + 1$