Adv Alegbra HW Solutions 144

Adv Alegbra HW Solutions 144 - 144 (ii) Show that one root...

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Unformatted text preview: 144 (ii) Show that one root of f ( X ) = X 3 + X 2 − 36 is an integer and find the other two roots. Compare your method with Cardano’s formula and with Vi` te’s trigonometric solution. e Solution. Corollary 3.91 says that any integer root is a divisor of 36, and one checks that 3 is a root; therefore, X 3 + X 2 − 36 = ( X − 3)( X 2 + 4 X + 12). √ The quadratic formula gives the other two roots: −4 ± 8, both of which are negative. The other two methods are longer. Both require the substitution X = x − 1 , yielding the reduced cubic x 3 − 1 x − 970 . We do not 3 3 27 give the calculations. 5.6 Show that if u is a root of a polynomial f (x ) ∈ R [x ], then the complex conjugate u is also a root of f (x ). Solution. Let f (x ) = ri xi . If 0 = ri u i , then since complex conjugation is an isomorphism fixing each real number, 0=0= ri u i = ri u i = f (u ). 5.7 Assume that 0 ≤ 3α < 360◦ . (i) If cos 3α is positive, show that there is an acute angle β with 3α = 3β or 3α = 3(β + 90◦ ), and that the sets of numbers cos β, cos(β + 120◦ ), cos(β + 240◦ ) and cos(β + 90◦ ), cos(β + 210◦ ), cos(β + 330◦ ) coincide. (ii) If cos 3α is negative, show that there is an acute angle β with 3α = 3(β + 30◦ ) or 3α = 3(β + 60◦ ), and that the sets of numbers cos(β + 30◦ ), cos(β + 150◦ ), cos(β + 270◦ ) cos(β + 60◦ ), cos(β + 180◦ ), cos(β + 270◦ ) and coincide. 5.8 Show that if cos 3θ = r , then the roots of 4x 3 − 3x − r are cos θ, cos(θ + 120◦ ), and cos(θ + 240◦ ). ...
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