144(ii)Show that one root off(X)=X3+X2−36 is an integer andfind the other two roots. Compare your method with Cardano’sformula and with Vi`ete’s trigonometric solution.Solution.Corollary 3.91 says that any integer root is a divisor of36, and one checks that 3 is a root; therefore,X3+X2−36=(X−3)(X2+4X+12).The quadratic formula gives the other two roots:−4±√8, bothof which are negative.The other two methods are longer. Both require the substitutionX=x−13, yielding the reduced cubicx3−13x−97027. We do notgive the calculations.5.6Show that ifuis a root of a polynomialf(x)∈R[x], then the complexconjugateuis also a root off(x).Solution.Letf(x)=∑rixi. If 0=∑riui, then since complex conju-gation is an isomorphismfixing each real number,0=0=riui=riui=f(u).5.7Assume that 0≤3α <360◦.(i)If cos 3
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