Factorize completely: \(x^{2} + x^{2}y + 3x - 10y + 3xy - 10\).

  • A (x + 2)(x + 5)(y + 1)
  • B (x + 2)(x - 5)(y + 1)
  • C (x - 2)(x + 5)(y + 1)
  • D (x - 2)(x - 5)(y + 1)
waec 2016furthermathematics

The correct answer is C. (x - 2)(x + 5)(y + 1)

\(x^{2} + x^{2}y + 3x - 10y + 3xy -10\)

= \(x^{2} + x^{2}y + 3x + 3xy - 10y - 10  = x^{2}(1 + y) + 3x(1 + y) - 10(y + 1)\)

= \((x^{2} + 3x - 10)(y + 1)\)

= \((x^{2} + 3x - 10) = x^{2} - 2x + 5x - 10\)

= \(x(x - 2) + 5(x - 2) = (x - 2)(x +5)\)

\(\therefore x^{2} + x^{2}y + 3x - 10y + 3xy -10 = (x - 2)(x + 5)(y + 1)\).

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