SELECTED CHAPTERS FROM ALGEBRA I. R. Shafarevich Preface

20 I. R. Shafarevichjxj n;k > nja kjja n j , i.e.,(18) jxj > n;k sn ja kjja n j :Therefore, it sis enough to choose for N an arbitrary number larger than all thenumbers n;k n ja kj, k =0 1 ...n; 1, and it will satisfy the assertion of Theorem5.ja n jTheorem 5 has a lot of useful corollaries. Note rst that under the assumptionsof the theorem (i.e., for jxj >N) we always have jf(x)j > 0, which followsimmediately from inequality (12)jf(x)j = ja 0 + a 1 x + + a n x n j 6 ja n x n j;ja 1 + a 1 x + + a n;1x n;1 j:But this means that the polynomial f(x) does not have roots x with jxj > N.In other words, roots of a polynomial (if they exist) have to be contained in thesegment jxj 6 N, where, as weshave shown (inequality (18)) N can be chosen asthe greatest of the numbers n;k n ja kj. One calls such anumber N the bound ofja n jroots of the polynomial. So, for the polynomial x 3 ; 7x + 5 one can take forN anarbitrary number greater than 3p 3 5 and p 3 7. For example, N =46 satisesthe conditions. This means that all roots of the polynomial are distributed between;46 and46. We have convinced ourselves earlier that they are in fact containedbetween ;3 and +3 (Table 1).Theorem 5 implies more that just the assertion that f(x) 6= 0ifjxj >N,forthe found value of N. To evaluate the value of a 0 + a 1 x + + a n;1x n;1 + a n x nmeans to sum up two real numbers a 0 + a 1 x + + a n;1x n;1 and a n x n , rst ofwhich is smaller (by absolute value) than the other (for jxj >N). But then thesign is determined by the sign of the second summand. We come to the followingconclusion:COROLLARY 1. For jxj >N,where N is the bound of roots dened inTheorem5, values of the polynomial f(x) have the same sign as the leading term a n x n .Suppose that the degree n of the polynomial is odd. Then the sign of theleading term a n x n for x > 0 agrees with the sign of the coecient a n , and forxN and x