WAEC Further Mathematics 2016 Paper

Concept:Factor the polynomial by grouping terms and then factoring the resulting quadratic expression.Explanation:Rewrite the expression by regrouping terms that contain common factors.$x^{2} + x^{2}y + 3x + 3xy - 10y - 10$Group as $x^{2}(1 + y) + 3x(1 + y) - 10(y + 1)$.Since $y + 1 = 1 + y$, factor out $(y + 1)$:$(y + 1)(x^{2} + 3x - 10)$.Now factor the quadratic $x^{2} + 3x - 10$.Find two numbers whose product is $-10$ and sum is $3$: these are $5$ and $-2$.So $x^{2} + 3x - 10 = (x + 5)(x - 2)$.Therefore, the complete factorisation is:$x^{2} + x^{2}y + 3x - 10y + 3xy - 10 = (x + 5)(x - 2)(y + 1)$.Answer:$(x - 2)(x + 5)(y + 1)$, which is option C.