Complex numbers and polynomials - University College Cork

Hence q ≤ 0 and27r 2 + 8q 3 ≤ 0are necessary conditions for the cubic to have three real roots.(This is an algebraic approach. What does a geometric approach suggest?)5. Are these conditions sufficient In other words, if they hold, will x 3 +qx+r havethree real roots? (Answer: not necessarily; consider 2x 3 − 3x + 1.) Supposeq ≤ 0 to begin with, and see what effect this has on the shape of the graphof y = x 3 + qx + r. (It will help you to visualise the shape of such a cubic.)The graph has two turning points given by x = ± √ −q/3. The correspondingy values are± 2q√−q3 3 + r.The smallest of these is √2q −q3 3 + r.The graph tells us that the cubic has three real roots if this is ≤ 0. This is soif27r 2 ≤ −4q 3 .In other words, the cubic has three real roots if 27r 2 + 4q 3 ≤ 0. (This impliesthat q < 0.)This, then, is a sufficient condition, which implies the necessary conditionobtained above, but the two conditions are different! Can the gap be closedbetween these two conditions? Is it the case, perhaps, that 27r 2 + 4q 3 ≤ 0 istrue if the roots are real? To see that this is indeed the case, we revisit ourattempt to derive a necessary condition assuming the roots are real. This time27r 2 + 4q 3 = 27a 2 b 2 c 2 + 4(ab + bc + ca) 3= 27a 2 b 2 (a + b) 2 − 4(a 2 + ab + b 2 ) 3= 3a 4 b 2 + 26a 3 b 3 + 3a 2 b 4 − 4a 6 − 12a 5 b − 4b 6 − 12b 5 a= −(a − b) 2 (a + 2b) 2 (b + 2a) 2≤ 0.(Query: how did we factorize this expression? By the way, this also points upthat the AM-GM inequality can sometimes be a blunt instrument.)So, to sum up: the cubic x 3 + qx + r has three real roots iff 27r 2 + 4q 3 ≤ 0. Ithas one real root and two complex roots iff 27r 2 + 4q 3 > 0.6. Here’s an alternative method of solving x 3 + qx + r = 0 that relies on thetrigonometric identity:4 cos 3 θ = cos(3θ) + 3 cos θ.28