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JAMB 1988 · UME · Q24

Evaluate 4a2−49b22a2+5ab−7b2\dfrac{4a^2 - 49b^2}{2a^2 + 5ab - 7b^2}.

Worked solution (try it first)
  1. The top is a difference of two squares: 4a2−49b2=(2a−7b)(2a+7b)4a^2 - 49b^2 = (2a - 7b)(2a + 7b).
  2. Factorise the bottom: 2a2+5ab−7b2=2a2−2ab+7ab−7b22a^2 + 5ab - 7b^2 = 2a^2 - 2ab + 7ab - 7b^2, which is (2a+7b)(a−b)(2a + 7b)(a - b).
  3. Cancel the common factor 2a+7b2a + 7b.
  4. This leaves 2a−7ba−b\dfrac{2a - 7b}{a - b}, option D.

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