JAMB 1990 · UME · Q2

The H.C.F. of a2bx+abx2a^2bx + abx^2 and a2b−b3a^2b - b^3 is

Worked solution (try it first)
  1. Factorise the first: a2bx+abx2=abx(a+x)a^2bx + abx^2 = abx(a + x).
  2. Factorise the second: a2b−b3=b(a2−b2)a^2b - b^3 = b(a^2 - b^2)
    =b(a−b)(a+b)= b(a - b)(a + b).
  3. The only factor in both is bb, so the H.C.F. is bb, option A.

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