JAMB 1989 · UME · Q21

Quadratics & their graphs

Factorize $4a^2 + 12ab - c^2 + 9b^2$ .

Worked solution (try it first)

Group the first, second and last terms: $4a^2 + 12ab + 9b^2 = (2a + 3b)^2$ .
So the expression is $(2a + 3b)^2 - c^2$ , a difference of two squares.
So it factorises as $(2a + 3b - c)(2a + 3b + c)$ , option B.

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