Past papers › JAMB 2018 · UTME › Question 13 Question JAMB General Maths 2018 Objective Quadratics & their graphs Expressions, formulae & change of subject Quadratics & their graphs, Expressions, formulae & change of subject
Simplify x − 7 x 2 − 9 × x 2 − 3 x x 2 − 49 \dfrac{x - 7}{x^2 - 9} \times \dfrac{x^2 - 3x}{x^2 - 49} x 2 − 9 x − 7 × x 2 − 49 x 2 − 3 x
A x ( x − 3 ) ( x + 7 ) \dfrac{x}{(x - 3)(x + 7)} ( x − 3 ) ( x + 7 ) x B ( x + 3 ) ( x + 7 ) x \dfrac{(x + 3)(x + 7)}{x} x ( x + 3 ) ( x + 7 ) C x ( x − 3 ) ( x − 7 ) \dfrac{x}{(x - 3)(x - 7)} ( x − 3 ) ( x − 7 ) x D x ( x + 3 ) ( x + 7 ) \dfrac{x}{(x + 3)(x + 7)} ( x + 3 ) ( x + 7 ) x
Worked solution (try it first) Factorise each part:
x 2 − 9 = ( x − 3 ) ( x + 3 ) x^2 - 9 = (x - 3)(x + 3) x 2 − 9 = ( x − 3 ) ( x + 3 ) ,
x 2 − 3 x = x ( x − 3 ) x^2 - 3x = x(x - 3) x 2 − 3 x = x ( x − 3 ) and
x 2 − 49 = ( x − 7 ) ( x + 7 ) x^2 - 49 = (x - 7)(x + 7) x 2 − 49 = ( x − 7 ) ( x + 7 ) .
So the product is
x − 7 ( x − 3 ) ( x + 3 ) × x ( x − 3 ) ( x − 7 ) ( x + 7 ) \dfrac{x - 7}{(x - 3)(x + 3)} \times \dfrac{x(x - 3)}{(x - 7)(x + 7)} ( x − 3 ) ( x + 3 ) x − 7 × ( x − 7 ) ( x + 7 ) x ( x − 3 ) .
Cancel the common factors
x − 7 x - 7 x − 7 and
x − 3 x - 3 x − 3 , top and bottom.
This leaves
x ( x + 3 ) ( x + 7 ) \dfrac{x}{(x + 3)(x + 7)} ( x + 3 ) ( x + 7 ) x , option D.
Watch out
The factor that cancels with x ( x − 3 ) x(x - 3) x ( x − 3 ) x − 3 x - 3 x − 3 x 2 − 9 x^2 - 9 x 2 − 9 x + 3 x + 3 x + 3 x − 3 x - 3 x − 3 Also set as JAMB 1983 · UME · Q15
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