WAEC - Further Mathematics (2016 - No. 7)
Factorize completely: \(x^{2} + x^{2}y + 3x - 10y + 3xy - 10\).
(x + 2)(x + 5)(y + 1)
(x + 2)(x - 5)(y + 1)
(x - 2)(x + 5)(y + 1)
(x - 2)(x - 5)(y + 1)
Explanation
\(x^{2} + x^{2}y + 3x - 10y + 3xy -10\)
= \(x^{2} + x^{2}y + 3x + 3xy - 10y - 10 = x^{2}(1 + y) + 3x(1 + y) - 10(y + 1)\)
= \((x^{2} + 3x - 10)(y + 1)\)
= \((x^{2} + 3x - 10) = x^{2} - 2x + 5x - 10\)
= \(x(x - 2) + 5(x - 2) = (x - 2)(x +5)\)
\(\therefore x^{2} + x^{2}y + 3x - 10y + 3xy -10 = (x - 2)(x + 5)(y + 1)\).
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