WAEC 2019 Mathematics · Question 22

Simplify: x2−5x−14x2−9x+14\frac{x^2 - 5x - 14}{x^2 - 9x + 14}

Options

  1. A.x−7x+7\frac{x - 7}{x + 7}
  2. B.x+7x−7\frac{x + 7}{x - 7}
  3. C.x−2x+4\frac{x - 2}{x + 4}
  4. D.x+2x−2\frac{x + 2}{x - 2}

Answer: D. x+2x−2\frac{x + 2}{x - 2}

Explanation written with AI and checked against the answer key

Step 1: Analyze the question We need to simplify a rational expression by factorising the numerator and denominator, then cancelling common factors.

Step 2: Concept Factorise each quadratic by finding two numbers whose product is the constant term and whose sum is the coefficient of x.

Step 3: Solution

Numerator: x² - 5x - 14

  • We need two numbers with product -14 and sum -5.
  • Those are -7 and +2 (since -7 × 2 = -14 and -7 + 2 = -5).
  • So x² - 5x - 14 = (x - 7)(x + 2)

Denominator: x² - 9x + 14

  • We need two numbers with product +14 and sum -9.
  • Those are -7 and -2 (since (-7)(-2) = 14 and -7 + -2 = -9).
  • So x² - 9x + 14 = (x - 7)(x - 2)

Put together:

x2−5x−14x2−9x+14=(x−7)(x+2)(x−7)(x−2)\frac{x^2 - 5x - 14}{x^2 - 9x + 14} = \frac{(x - 7)(x + 2)}{(x - 7)(x - 2)}

Cancel the common factor (x - 7), where x ≠ 7:

=x+2x−2= \frac{x + 2}{x - 2}

Step 4: Answer The simplified form is (x + 2)/(x - 2), which is option D.

Step 5: Exam tip Always factorise first before cancelling. Never cancel individual terms such as the x² or the numbers directly. Only whole factors can be cancelled. You can also check quickly by substituting a value, e.g. x = 3: the original gives (9 - 15 - 14)/(9 - 27 + 14) = (-20)/(-4) = 5, and option D gives 5/1 = 5. ✓

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