Given the equation x^3-12x^2+ax-48=0 has roots p, 2p and 3p, find p and a.

QUESTION: Given the equation x³-12x²+ax-48 = 0 has roots p, 2p and 3p, find p and a. Roots mean x = p, x = 2p and x = 3p hence (x-p), (x-2p) and (x-3p) are factors of the equation. Expanding these three factors together will equal the equation. (x-p)(x-2p)(x-3p) = (x²-px-2px+2p²)(x-3p) = (x³-px²-2px²+2p²x-3px²+3p²x+6p²x-6p³) = 0. By collecting and equating coefficients both p and a can be found. x³+(-p-2p-3p)x²+(2p²+3p²+6p²)x-6p³ = x³-6px²+11p²x-6p^{3 }-6p³ = -48 hence p³ = 8 and so p = 2 11p² = a and so a = 44 ANSWER: p = 2 and a = 44