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◦ ). ...
View
Full
Document
This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

Click to edit the document details