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-6p3 -6p3 = -48 hence p3 = 8 and so p = 2 11p2 = a and so a = 44 ANSWER: p = 2 and a = 44
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