Some Polynomial Theorems. John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA


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1 Some Polynomial Theorems by John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA
2 This paper contains a collection of 31 theorems, lemmas, and corollaries that help explain some fundamental properties of polynomials. The statements of all these theorems can be understood by students at the precalculus level, even though a few of these theorems do not appear in any precalculus text. However, to understand the proofs requires a much more substantial and more mature mathematical background, including proof by mathematical induction and some simple calculus. Of significance are the Division Algorithm and theorems about the sum and product of the roots, two theorems about the bounds of roots, a theorem about conjugates of irrational roots, a theorem about integer roots, a theorem about the equality of two polynomials, theorems related to the Euclidean Algorithm for finding the!"# of two polynomials, and theorems about the Partial Fraction Decomposition of a rational function and Descartes's Rule of Signs. It is rare to find proofs of either of these last two major theorems in any precalculus text. 1. The Division Algorithm If $%&' and (%&' )/ * are any two polynomials then there exist unique polynomials +%&' and,%&' such that $%&'  (%&'. +%&' /,%&' where the degree of,%&' is strictly less than the degree of (%&' when the degree of (%&' 0 1 or else,%&' ) *. Division Algorithm Proof: We apply induction on the degree 2 of $%&'3 We let 4 denote the degree of the divisor (%&'3 We will establish uniqueness after we establish the existence of +%&' and,%&'3 If 2* then $%&'5where 5is a constant3 Case 1: 4  *3 (%&'  6 where 6 is a constant and since (%&' ) 8 * we know In this case choose +%&'  and choose,%&' ) *3 6 5 Then (%&'. +%&' /,%&'  6. / * $%&'3 In this case,%&' ) *3 6 Case 9: 4 : *3 In this case let +%&' ) * and let,%&'  53 Then clearly (%&'. +%&' /,%&'  (%&'.*/5*/55$%&'3 In this case the degree of,%&' is strictly less than the degree of (%&'3 Now assume there exist polynomials + 1%&' and, 1%&' such that $ 1%&'  (%&'. + 1%&' /, 1%&' whenever $%&' 1 is any polynomial that has a degree less than or equal to 6. 6/1 6 Let $%&' be a polynomial of degree 6 / 13 We assume $%&'  ; + & / ; & / < / ; & / ; where ; 6/1 7 *3 We must show the theorem statement holds for $%&'3 Case 1: 4  *3 (%&'6where 6 is a constant and since (%&')* / we know 67*3 1 Let +%&'  $%&' and let,%&' ) *3 6 1 Then (%&'. +%&' /,%&'  6. $%&' / *  $%&' / *  $%&'3 In this case,%&' ) * * proof continued on the next page page 1
3 Case 9: 4 : *3 4 ; 6/1 Let (%&'( 4& /</(&/( 1 * where ( 4 7*3 Note that 7* since both ( 4 ; 6/1 6/1=4 constants are nonzero. Let $ 1%&'  $%&' = &. (%&'3 Then the subtraction on ( 4 the right cancels the leading term of $%&' so $ 1%&' is a polynomial of degree 6 or less and we can apply the induction assumption to $%&' 1 to conclude there exist polynomials + 1%&' and, 1%&' such that $ 1%&'  (%&'. + 1%&' /, 1%&' where the degree of, 1%&' is strictly less than that of (%&'3 ; 6/1 6/1=4 $ 1%&'  (%&'. + 1%&' /, 1%&'  $%&' = &. (%&' ( 4 Now we solve the 2nd equation for $%&'3 ; 6/1 6/1=4 $%&'  &. (%&' / (%&'. + 1%&' /, 1%&' ( 4 ; 6/1 6/1=4 $%&'  (%&'.! & / + 1%&' "/, 1%&'3 ( 4 ; 6/1 6/1=4 So we may let +%&' ! & / + 1%&' " and let,%&' , 1%&' and ( 4 we have established the theorem holds for $%&' of degree 6 / 13 The induction proof that establishes the existence part of the theorem is now complete. To establish uniqueness, suppose $%&'  (%&'. + 1%&' /, 1%&'  (%&'. + 9%&' /, 9%&'3 Then we have (%&'. # + %&' = + %&' $ , %&' =, %&'3 Call this equation (*) Case 1: 4*3 In this case both remainders must be identically zero and this means, 1%&' ), 9%&'3 In turn, this means (%&'. > + 1%&' = + 9%&'? ) *, and since (%&' ) 8 * we must have + %&' = + %&' ) * which of course implies + %&' ) + %&' Case 9: 4 : *3 If # +%&'=+%&' 1 9 $ )* 8 then we can compute the degrees of the polynomials on both sides of the (*) equation. The degree on the left side is greater than or equal to the degree of (%&'3 But on the right side, both remainders have degrees less than (%&' so their difference has a degree that is less than or equal to that of either which is less than the degree of (%&'3 This is a contradiction. So we must have # +%&'=+%&' 1 9 $ )* and when this is the case the entire left side of the (*) equation is identically * and we may add back, 1%&' from the right side to conclude that the two remainders are also identically equal. page 2
4 2. The Division Check for a Linear Divisor Consider dividing the polynomial $%&' by the linear term %& = ;'3 Then, the Division Check states that: $%&'  %& = ;'. +%&' /, Division Check Proof: This is just a special case of the Division Algorithm where the divisor is linear. 3. Remainder Theorem When any polynomial $%&' is divided by %& = ;' the remainder is $%;'3 Remainder Theorem Proof: By the Division Check we have $%&'  %& = ;'. +%&' /,3 Now let &  ;3 This last equation says $%;'  %; = ;'. +%;' /, $%;'*.+%;'/,*/,,3 4. Factor Theorem %& = ;' is a factor of the polynomial $%&' if and only if $%;'  *3 Factor Theorem Proof: Assume %& = ;' is a factor of $%&'. Then we know %& = ;' divides evenly into $%&'3 The remainder when $%&' is divided by %& = ;' must be 0. By the Remainder Theorem this says * ,  $%;'3 Next, assume $%;'  *3 Divide $%&' by %& = ;'3 By the Remainder Theorem, the remainder is $%;'  *3 Since the remainder is 0, the division comes out even so that %& = ;' is a factor of $%&'3 5. Maximum Number of Zeros Theorem A polynomial cannot have more real zeros than its degree. Maximum Number of Zeros Theorem Proof: By contradiction. Suppose $%&' has degree 201, and suppose /1 are 2+1 roots of $%&'3 By the Factor Theorem, since $%; 1'  * then there exists a polynomial + 1%&' of degree one less than $%&' such that $%&'  %& = ; 1'. + 1%&'3 Now since $ %; 9&  * and since ; 9 7 ; we must have +%;'* 1 9 and again by the Factor Theorem we can write $%&'%&=;'.%&=;'.+%&' where + 9%&' is of degree 9 less than $%&'3 Now since ; B is distinct from ; 1 and ; 9 we must have +%;'* 9 B and we can continue to factor $%&'%&=;'.%&=;'.%&=;'.+%&' 1 9 B B where the degree of + B%&' is of degree B less than $%&'3 Clearly this argument can be repeated until we reach the stage where $%&'%&=; 1'<%&=; 2'.+ 2%&' and + 2%&' is of degree 2 less than $%&'3 Since $%&' only had degree 2 in the first place, + 2%&' must be of degree 0 making + 2%&' some constant, say + 2%&'  53 Now ; 2/1 is still a zero of and since ; 2/1 is distinct from all the other C we must have + 2%; 2/1'*3 The only way this can happen is if 5* and this would imply a contradiction since we are assuming 2013 page 3
5 6. Fundamental Theorem of Algebra a) Every polynomial of degree 201has at least one zero among the complex numbers. b) If $%&' denotes a polynomial of degree then $%&' has exactly 2 roots, some of which may be either irrational numbers or complex numbers. Fundamental Theorem of Algebra Proof: This is not proved here. Gauss proved this in 1799 as his Ph.D. doctoral dissertation topic. 7. Product and Sum of the Roots Theorem 2 2=1 B 9 Let $%&' 1 & /; 2=1& /</; B& /; 9& /; 1 &/;* be any polynomial with real 2 coefficients with a leading coefficient of 1 where 201. Then ;* is %=1' times the product of all the roots of $%&'  * and ; 2=1 is the opposite of the sum of all the roots of $%&'  *. Product and Sum of the Roots Theorem Proof: By the Fundamental Theorem of Algebra we know $%&' has 2 roots which may be denoted 1 9 B 2 Now form the product of the 2 factors associated with these roots. Let +%&' %& =, 1'%& =, 9'%& =, B'<%& =, 2' and multiply out all these terms. Then inspect the coefficient on & 2=1 and inspect the constant term. This can also be formally proved by using induction on 2. When 21 we have $%&'&/; * and in this case the only root of $%&' is, 1  =; * 3 Since, 1 is the only root,, 1 is itself the product of all 2 1 the roots. But then %=1' ;*  %=1' ;*  =; * , 13 So this establishes the part about the constant term. Note again that since, 1 is the only root,, 1 is itself the sum of all the roots and the 2nd leading coefficient is the opposite of the sum of all the roots since ;*  =%, 1'3 It is probably more instructive to manually look at the case when 29before setting up the 9 induction step. Note that %& =, 1'%& =, 9'  & / %=, 1 / =, 9'& /, 1, 93 In this case it is immediately apparent that the 2nd leading coefficient is the opposite of the sum of all the roots and the constant term is product of all the roots. Because $%&' is quadratic, in this case 29 so 2 9 ( = 1)  ( = 1)  13 Now lets assume the result is true whenever we have 6 roots and let $%&' be a polynomial with 6 / 1 roots, say $%&'  %& =, 1'%& =, 9'<%& =, 6/1'3 Now consider that we may write $%&'  '%& =, 1'%& =, 9'<%& =, 6' (%& =, 6/1'. Let +%&'  '%& =, 1'%& =, 9'<%& =, 6' ( 3 Then +%&' has degree 6 and we may apply the induction hypothesis to +%&'. If we write 6 6 6=1 +%&'& /; 6=1& /</; 1 &/;* then we know ; 6=1 =) +, C* and we know 6 ;*  %=1'.),, C*. 6 C1 Now $%&'+%&'.%&=, '%&=, '. 6 6=1 & /; & /</; &/;. C1 6/1 6/1 6=1 1 * proof continued on the next page page 4
6  6/1 6 9 & /; 6=1& /</; 1 & /;*&./  6 6=1 %=, '& / %=, '; & / < / %=, '; & / %=, ';. 6/1 6/1 6=1 6/1 1 6/1 * 6/1 6 & /D; /%=, 'E& /</%; =,.; '&/%=, '.; 6=1 6/1 * 6/1 1 6/1 *. Clearly 6 6/1 ; 6=1 / %=, 6/1'  = ) +, 6* / %=, 6/1'  = ) +, C* and %=, 6/1 '. ;*  C1 C1 6 6/1 6 6/1 %=, 6/1'. %=1'. ),, C*  %=1'. ),, C* which are what we needed to establish. C1 C1 8. Rational Roots Theorem 2 2=1 B 9 Let $%&'; 2& /; 2=1& /</; B& /; 9& /; 1 &/;* be any polynomial 5 with integer coefficients. If the rational number is a root of $%&'  * then 5 must be a factor ( of ;* and ( must be a factor of ; 23 Rational Roots Theorem Proof. ; 2=1 2=1 ; 2=9 2=9 ; B B ; 9 9 ; 1 ;* Let +%&'  & 2 / & / & /</ & / & / &/ 3 ; 2 ; 2 ; 2 ; 2 ; 2 ; 2 By the Product of the Roots Theorem, we know the product of the roots of this 2 polynomial is the fraction %=1'. ; * 3 Thus if 5 is a root, 5 must be a factor of ;* and ( must ; 2 ( be a factor of ; Integer Roots Theorem 2 2=1 B 9 Let $%&'& /; 2=1& /</; B& /; 9& /; 1 &/;* be any polynomial with integer coefficients and with a leading coefficient of 1. If $%&' has any rational zeros, then those zeros must all be integers. Integer Roots Theorem Proof: By the Rational Roots Theorem we know the denominator of any rational zero must divide into the leading coefficient which in this case is 1. Thus any denominator must be F 1 making the rational zero into a pure integer. page 5
7 10. Upper and Lower Bounds Theorem Let $%&' be any polynomial with real coefficients and a positive leading coefficient. ( Upper Bound ) If ; : * and $%;' : * and if in applying synthetic substitution to compute $%;' all numbers in the 3rd row are positive, then ; is an upper bound for all the roots of $%&'  *. ( Lower Bound ) If ; G * and $%;' 7 * and if in applying synthetic substitution to compute $%;' all the numbers in the 3rd row alternate in sign then ; is a lower bound for all the roots of $%&'  *. [ In either bound case, we can allow any number of zeros in any positions in the 3rd row except in the first and last positions. The first number is assumed to be positive and the last number is $%;' 7 *. For upper bounds, we can state alternatively and more precisely that no negatives are allowed in the 3rd row. In the lower bound case the alternating sign requirement is not strict either, as any 0 value can assume either sign as required. In practice you may rarely see any zeros in the 3rd row. However, a slightly stronger and more precise statement is that the bounds still hold even when zeros are present anywhere as interior entries in the 3rd row.] Upper and Lower Bounds Theorem Proof: (Upper Bound). Let H be any root of the equation $%&'  *3 Must show H G ;3 If H*, then clearly HG; since ; is positive in this case. So we assume H7*. If the constant term of $%&' is *, then we could factor & or a pure power of & from $%&' and just operate on the resulting polynomial that is then guaranteed to have a nonzero constant term. So we can implicitly assume $%*' 7 *3 The last number in the third row of the synthetic substitution process is positive and it is $%;'3 Since H is a root, we know by the Factor Theorem that $%&'  %& = H'. +%&' where +%&' is the quotient polynomial. The leading coefficient of $%&' is also the leading coefficient of +%&' and since all of +%&' 's remaining coefficients are positive, and since ; : *, we must have +%;' : *3 Finally, $%;'  %; = H'. +%;'3 Since +%;' : *, we may $%;' divide by +%;' and get %; = H'  3 Now since $%;' and +%;' are both positive, %; = H' : * +%;' which implies HG;3 Note that since the leading coefficient of +%&' is positive and since ;:*, we don't really need all positive numbers in the last row. As long as +%&' 's remaining coefficients are nonnegative we can guarantee that +%;' : *3 (Lower Bound). Let H be any root of the equation $%&'  *3 Must show ; G H3 As in the above Upper Bound proof, we can easily dispense with the case when H*3 Clearly ;GHwhen H* because ; is negative. We can further implicitly assume no pure power of & is a factor of $%&' and this also allows us to assume $%*' 7 *3 Since $%H'  * by the Factor Theorem we may write $%&'%&=H'.+%&'. Substituting &; we have $%;'%;=H'.+%;'3 Since $%;' $%;' 7 * we know +%;' 7 *3 So we can divide by +%;' to get %; = H'  3 +%;' Now +%;' is either positive or negative. Because ; G * and the leading term in +%&' has a positive coefficient, the constant term in +%&' has the same sign as +%;'. This fact can be established by considering the two cases of the even or odd degrees that +%&' must have. proof continued on the next page page 6
8 For examples: J I B 9 +%&'  1& = 9& / B& = I& / J& = K3 With ; G +%;' G * and +%;' and +%&' L s constant term agree in sign. or I B 9 +%&'  1& =9& /B& =I&/J. With ; G *, +%;' : * and again +%;' and +%&' L s constant term agree in sign. We might note that in these examples, it would make no difference if any of the interior coefficients were 0. This is because the first term has a positive coefficient, and all the remaining terms just add fuel to the fire with the same sign as the first term. The presence of an interior zero just means you might not get as big a fire, but the first term guarantees there is a flame! Another note is that $%*'  %=H'. and since we are assuming H 7 we can divide this equation by =H to conclude that +%*' 7 * when $%*' 7 *3 So assuming neither H nor the constant term in $%&' are zero guarantees that the constant term in +%&' must be strictly positive or strictly negative. Since the numbers in the third row alternate in sign, $%;' differs in sign from the constant term in +%&'3 But since the constant term in +%&' has the same sign as +%;' we know $%;' and +%;' $%;' differ in sign. So %;=H' G*3 ;GH3 +%;' 11. Intermediate Value Theorem If $%&' is any polynomial with real coefficients, and if $%;' : * and $%H' G * then there is at least one real number 5 between ; and H such that $%5'  *3 Intermediate Value Theorem Proof: This result depends on the continuity of all polynomials and is a special case of the Intermediate Value Theorem that normally appears in a calculus class. 12. Single Bound Theorem 2 2=1 2=9 B 9 Let $%&'& /; 2=1& /; 2=9& /</; B& /; 9& /; 1 &/;* be any polynomial with real coefficients and a leading coefficient of 1. Let M11/4;&D 9/ 2=1/ E and let /;*///; 1/ //; 9/ /A //; 2=1/ E. Finally let M  4C2DM Then every zero of $%&' lies between =M and M. Single Bound Theorem Proof: We need to show M is an Upper Bound and we need to show =M is a Lower Bound. Case 1: MM3 1 Then we know for *NCN2=1that M01/ /; C/ 3This implies two things. First, M01and second, M = /; C/ 0 13 These two inequalities are crucial and further imply that =M N =1 and =M / /; / N =13 C proof continued on the next page page 7
9 To show M is an Upper Bound, consider the synthetic substitution calculation of $%M'3 We will label the second and remaining coefficients in the second row as H C values. We will label the second and remaining coefficients in the third row as values. 5 C 1 ; ; ; < ; ; M H H < H H < 5 5 2=1 2=9 2=B 1 * 2=9 2=B 1 * 2=1 2=9 2=B 1 * M We claim that each 5C value is not only positive, we claim each 5C 0 13 Similarly we claim each HC 0 M3 We will establish these two claims by working from left to right across the columns in the synthetic substitution table, one column at a time. First note that 52=1 ; 2=1 /M0=; / 2=1/ /MM=/; 2=1/ 013 We are done with the 2nd column. Now we will argue about the 3rd column in the above table. Having established in the 2nd column that 52=1 0 multiply both sides of this inequality by M to obtain: H2=952=1.M 0 M3 Now we basically repeat the above argument to establish the size of 52=9  ; 2=9 /H2=93 Here we use the fact that H 0 M 0 1/ /; /. So H = /; / =9 2=9 2=9 2=9 So 52=9 ; 2=9 /H2=9 0=; / 2=9/ /H2=9 H 2=9 = /; 2=9/ 013 We are now done with the 3rd column in the table. Each next column is handled like the 3rd column. Just to make sure you get the idea we will establish our claims for the 4th column. Since 52=9 0 1, we can multiply across this inequality by M to get 52=9.M 0 M3 H2=B  52=9.M 0 M3 52=B ; 2=B /H2=B 0=; / 2=B/ /H2=B H 2=B = /; 2=B/ 0M= /; 2=B/ 013 Clearly we can continue working across the columns of the above table, one column at a time. Since all the 5 coefficients are positive we know M is an Upper Bound for the zeros of $%&'3 Next, to show C =M is a Lower Bound, consider the synthetic substitution calculation of $%=M'3 1 ; ; a < ; ; =M H H < H H < 5 5 2=1 2=9 2=B 1 * 2=9 2=B 1 * 2=1 2=9 2=B 1 * =M proof continued on the next page page 8
10 We claim that the coefficients in the 3rd row alternate in sign. Obviously the first coefficient is 1:*3 We claim not only that the 5C alternate in sign, we claim that when 5C G * then 5C N =13 We also claim that when 5 : * then C C In the 2nd column of the table we have 52=1 ; 2=1 =MN /; 2=1/ =M=M/; / 2=1/ N=13 So we have established our claim within the 2nd column. Moving over to the 3rd column we note that H2=952=1.% =M& and since both of the numbers in this product are negative, we have H 2=9 : *3 In fact when we start with the inequality that 52=1 N =1 and multiply across by the negative number =M we get 52=9. % =M& 0 M3 But H2=952=9.% =M& so we know that H2=9 0 M3 Now lets compute 52=93 52=9 ; 2=9 /H2=9 0; 2=9 /M0=; / 2=9/ /MM=/; 2=9/ 013 Next, consider what happens in the 4th column of the above table. We just established that 52= Multiplying across this inequality by =M we get 52=9. % =M& N =M3 But H2=B  52=9. % =M& so we know H2=B N =M3 Now lets compute 52=B3 52=B  ; 2=B /H2=B N ; 2=B =M N /; 2=B/ =M  =M //; 2=B/ N =13 Clearly the above arguments may be repeated as we move across the columns of the above table. Each time we multiply by =M to compute the next H C value we have a sign change. This is primarily why the values alternate in sign. 5 C In any case, the values in the 3rd row alternate in sign and since =M G * we know =M is a lower bound for any zero of the equation $%&'  *3 Case 2:! "! # 3 Subcase 1: M  1 and /; ///; ///; //<//; / G 13 * 1 9 2=1 In particular, for each C where *NCN2=1 we know *N/; C/ G1M3 Then = /; C/ /M1=/; C/ :*3 Since M1, the Synthetic Substitution table takes on a particularly simple form. Note how the 2nd row elements are the same as the 3rd row elements shifted over one column. 1 1 ; 2=1 ; 2=9 a2=b ; 2=I < ; 1 ;* 1 52=1 52=9 52=B < =1 52=9 52=B 52=I < 51 5* We claim that all the 5 C values are positive. Starting in the 2nd column, 52=1 ; 2=1 /10=; / 2=1/ /11=/; 2=1/ :*3 proof continued on the next page page 9
11 Now consider the 3rd column. 52=9  ; 2=9 /52=1  ; 2=9 /; 2=1 /1 0 =; / //=; / //1=%/; ///; /&/1:*3 2=9 2=1 2=9 2=1 Next consider the 4th column. c 2=B ; 2=B /5 2=9 ; 2=B /; 2=9 /; 2=1 /10 =; / //=; / //=; / //1=%/; ///; ///; /&/1:*3 2=B 2=9 2=1 2=B 2=9 2=1 Clearly we can continue to accumulate the sums of more and more terms and still apply the main inequality that appears in the Subcase 1: statement. So all the elements in the last row are positive and 1M is an upper bound for all the roots of $%&'* by applying the Upper/Lower Bounds Theorem. To establish that = 1 is a lower bound we compute synthetic substitution with =M  =13 1 ; ; a ; < ; ; =1 =5 =5 =5 < =5 = < 5 5 2=1 2=9 2=B 2=I 1 * 2=1 2=9 2=B 9 1 2=1 2=9 2=B 2=I 1 * =1 Now we must establish that the 5 C values in the last row alternate in sign. Starting in the 2nd column, 52=1  ; 2=1 / %=1' N /; 2=1/ / % =1& G *3 In the 3rd column, 52=9 ; 2=9 /% =52=1& ; 2=9 =; 2=1 /10 =; / //=; / //1=%/; ///; /&/1:*3 2=9 2=1 2=9 2=1 In the 4th column, 52=B ; 2=B /% =52=9& ; 2=B =; 2=9 /; 2=1 =1N /; ///; ///; / =1 G *3 2=B 2=9 2=1 Clearly this argument may be repeated to establish that the coefficients in the 3rd row really do alternate in sign. So M=1is a lower bound by the Upper/Lower Bounds Theorem. Subcase #: M  /; ///; ///; //<//; / and M 0 13 * 1 9 2=1 To establish that M is an upper bound, we consider the synthetic substitution table for computing $%M' and we will show that all the values in the last row are nonnegative. 1 ; ; ; < ; ; M H H < H H < 5 5 2=1 2=9 2=B 1 * 2=9 2=B 1 * 2=1 2=9 2=B 1 * M proof continued on the next page page 10
12 Now consider the 2nd column in the above table. 52=1  ; 2=1 /M 0 = /; 2=1 //M  /;*///; 1/ //; 9/ /<//; 2=9/ 0 *3 Next consider the 3rd column in the above table. Since 2=1.M052=13 But H2=952=1.Mso H2=9 052=13 Next, 52=9  ; 2=9 /H2=9 0 ; 2=9 /52=1  ; 2=9 /; 2=1 /M 0 = /; //=/; //M  /; ///; ///; //<//; / 0 *3 2=9 2=1 * 1 9 2=B We continue to argue in the same manner for the 4th column. Since 2= 2.M052= 23But H2= 352= 2.Mso H2= 3 052= 23 Next, 52= 3  ; 2=B /H2=B 0 ; 2=B /52=9 0 ; 2=B /; 2=9 /; 2=1 /M 0 =; / //=; / //=; / //M /; ///; ///; //<//; / 0*3 2=B 2=9 2=1 * 1 9 2=I This argument may be repeated across the columns in the above table to establish that all the numbers in the last row are nonnegative. So by the Upper/Lower Bounds Theorem, M is an upper bound for all the roots of $%&'  *3 Finally, consider the synthetic substitution table for computing $%=M'3 1 ; ; ; ; < ; ; =M H H H < H H < 5 5 2=1 2=9 2=B 2=I 1 * 2=9 2=B 2=I 1 * 2=1 2=9 2=B 2=I 1 * =M We claim the nonnegative numbers in the 3rd row of this table alternate in sign. In the first column the number is 1 so we know we are starting with a positive value. Now look at 52=1 in the 2nd column3 52=1 ; 2=1 /% =M& ; 2=1 /' =; / *//=; / 1/ /=; / 9/ /</=; / 2=1/ (  ; 2=1 /=; / 2=1 / /' =; / *//=; / 1/ /=; / 9/ /</=; / 2=9/ ( 3 Now the last term in the above expression is obviously less than or equal to zero, and the first two terms either make * or make =9. /; 2=1/ so the whole expression is less than or equal to 0. Now consider the 3rd column. We must show 52=9 0 *. We take the worst case from 5 2=1 assuming this has the smallest absolute value where 52=1 ' =; / *//=; / 1/ /=; / 9/ /</=; / 2=9/ ( 3 Then 52=9 ; 2=9 /' =; / *//=; / 1/ /=; / 9/ /</=; / 2=9/ (.% =M&  ; 2=9 /M./; 2=9 / /' =; / *//=; / 1/ /=; / 9/ /</=; / 2=B/ (.% =M& 0 ; 2=9 //; 2=9 //' =; / *//=; / 1/ /=; / 9/ /</=; / 2=B/ (.% =M& 0* since the third term is nonnegative and the first two terms make either 0 or 9. /; / 3 2=9 proof continued on the next page page 11
13 Now consider the 4th column. We must show 52=B N *3 We take the worst case from 52=9 assuming 52=9 has the smallest absolute value where 52=9 ' =; / *//=; / 1/ /=; / 9/ /</=; / 2=B/ (.% =M& 3 9 Then 52=B ; 2=B /' =; / *//=; / 1/ /=; / 9/ /</=; / 2=B/ (.% M &  ; 2=B =M 9./; 2=B //' =; / *//=; / 1/ /=; / 9/ /</=; / 2=I/ (.% M 9 & N 9 ; 2=B = /; 2=B //' =; / *//=; / 1/ /=; / 9/ /</=; / 2=I/ (.% M & N* since the third term is negative and the first two terms make either * or =9. /; / 3 Just to make sure you get the idea we will continue with the 5th column. We must show 52=I 0 *3 We take the worst case from 52=B assuming 52=B has the smallest absolute 9 value where 52=B ' =; / *//=; / 1/ /=; / 9/ /</=; / 2=I/ (.% M & 3 B Then 52=I ; 2=I /' =; / *//=; / 1/ /=; / 9/ /</=; / 2=I/ (.% =M &  B B ; 2=I /M./; 2=I //' =; / *//=; / 1/ /=; / 9/ /</=; / 2=J/ (.% =M & 0 B ; 2=I //; 2=I //' =; / *//=; / 1/ /=; / 9/ /</=; / 2=J/ (.% =M & 0* since the third term is nonnegative and the first two terms make either 0 or 9. /; / 3 This argument may be repeated across the columns of the above table to conclude that the nonnegative terms in the last row alternate in sign. By the Upper/Lower Bounds Theorem we know =M is a lower bound for all the zeros of $%&'  *3 2=B 2=I 13. Odd Degree Real Root Theorem If $%&' has real coefficients and has a degree that is odd then it has at least one real root. Odd Degree Real Root Theorem Proof: Without loss of generality we assume the leading coefficient of $%&' is positive. Otherwise we can factor =1 from $%&' apply the theorem to the polynomial that is the other factor. By choosing M : * sufficiently large we can establish that $%M' : * and $%=M' G *3 For example, see the above Single Bound Theorem. Now apply the Intermediate Value Theorem. There exists a number ; such that =M G ; G M and $%;'  *3 ; is real and is a root of $%&'3 page 12
14 14. Complex Conjugate Roots Theorem If $%&' is any polynomial with real coefficients, and if ; / HC is a complex root of the equation $%&'  *, then another complex root is its conjugate ; = HC3 (Complex number roots appear in conjugate pairs) Complex Conjugate Roots Theorem. This proof just depends on properties of the conjugate operator denoted by bars below. 2 2=1 B 9 If $%&'; 2& /; 2=1& /</; B& /; 9& /; 1 &/;* * then ; & 2 /; & 2=1 /</; & B 2 2=1 B /; 9& 9 /; 1 &/;* *. ; & 2 /; & 2=1 /</; & B 2 2=1 B /; 9& 9 /; 1 &/;* *. 2 2=1 B ; 2& /; 2=1& /</; B& /; 9&/; 1 &/;* *. 2 2=1 B ; 2& /; 2=1& /</; B& /; 9&/; 1 &/;* *. This shows $%&'*3 15. Linear and Irreducible Quadratic Factors Theorem Any polynomial $%&' with real coefficients may be written as a product of linear factors and irreducible quadratic factors. The sum of all the degrees of these component factors is the degree of $%&'3 Linear and Irreducible Quadratic Factors Theorem Proof: By the Fundamental Theorem of Algebra, $%&'  5%& =, 1'%& =, 9'%& =, B'<%& =, 2' where the, 6 denote the 2 roots of $%&'3 The constant 5 is simply $%&' 's leading coefficient. If all the, 6 roots are real then $%&' is a product of linear real factors only. However, if any, 6 value is a complex number then it can be paired with some other, O value which is its complex conjugate. If we assume, 6 ;/HC then, O ;=HC3 Note that since, 6 is complex, we must have H7*3 9 9 Next we note that, 6 /, O  9; and, 6., O  ; /H and we compute (& =, 6'%& =, O'  & = %, 6 /, O'& /, 6, O  & = %9;'& / %; / H '3 That this last expression is an irreducible quadratic factor follows by computing its discriminant that is %=9;' = I. 1. %; / H '  I; = I; = IH  =IH G * since H 7 *3 If we reorder or rename the indices of the roots so that, 6 and, O were the last two roots then we can assume that $%&' now takes the form in which we put the irreducible quadratic just found at the end $%&'  5%& =, 1'%& =, 9'<%& =, 2=B' '%& = 9;& / ; / H '( 3 Now if any two of the preceding linear factors form complex,value conjugate roots, we treat them just like we did with the pair, 6 and, O. This produces another irreducible quadratic factor which we also place as the last rightmost factor. Clearly this process can be continued until only real linear factors remain at the beginning and only irreducible quadratic factors are at the end. There may be no linear factors at the beginning and only irreducible quadratic factors, or there may be no irreducible quadratic factors at the end and only linear factors at the beginning. It all depends on the nature of the roots, C and how many of these roots are real and how many are complex. page 13
15 16. Irrational Conjugate Roots Theorem Let $%&' be any polynomial with rational real coefficients. If ; / H05 is a root of the equation $%&'  * where 05 is irrational and ; and H are rational, then another root is ; = H05 3 (Like complex roots, irrational real roots appear in conjugate pairs, but only when the polynomial has rational coefficients.) Irrational Conjugate Roots Theorem Proof: Assume ;/H05 is one root. Must show ;=H05 is also a root. If H* we are done, so assume H7*3 Let P%&' 1&= 1;/H &= 1;=H0522 %&=;& 9 9 =H 53 Then P%&' is a quadratic polynomial with rational coefficients. Next, consider dividing $%&' by P%&'3 By the Division Algorithm, there is a quotient polynomial +%&' and there exists a remainder polynomial,%&' such that $%&'  P%&'. +%&' /,%&' where the degree of,%&' is 1 or 0. If we assume,%&'  "& / # then " and # must be rational. In fact, since $%&' and P%&' have only rational coefficients, both +%&' and,%&' must have only rational coefficients. So we may write $%&'  P%&'. +%&' / "& / # and when we substitute &  ; / H05 we conclude that *" 1;/H052/# from which we can further conclude that "#*3 So we really have $%&'  P%&'. +%&'3 Finally we substitute &  ; = H05 in this last equation to conclude that $;=H * which is what we needed to show. 17. Descartes's Rule of Signs Lemma 1. If $%&' has real coefficients, and if $%;'  * where ; : then $%&' has at least one more sign variation than the quotient polynomial +%&' has sign variations where $%&'  %& = ;'+%&'3 [When the difference in the number of sign variations is greater than 1, the difference is always an odd number.] Descartes's Rule of Signs Lemma 1 Proof: The following particular example shows that +%&' may indeed have fewer sign variations than $%&' has. In this example, $%&' has three variations in sign while +%&' has only two variations in sign. Had the number / 152 in the top row been / 102 instead, then +%&' would have had even fewer sign variations as is shown in the next example =QQ 1J9 =QQ =B* 9I =1*K R9 B* 19 =JB IK 1J * In the next example, $%&' has four sign variations while +%&' has only one sign variation. So the difference of the sign variation counts in the example below is the odd number B =QQ 1*9 =QQ 1Q* 9I =1*K =S =1Q* 19 =JB =I =SJ * proof continued on the next page page 14
16 Finally we show one more example before starting the formal proof. In this example, $%&' has four sign variations and +%&' has only three sign variations. Moreover, the coefficients in $%&' and +%&' match signs column by column from left to right through the constant column in +%&'3 B 1 =K 11 =11 1J B =R K =1J 1 =B 9 =J * Assume the leading coefficient of $%&' is positive and consider the synthetic substitution form used to compute $%;'3 Consider the constant term in $%&'3 If this constant term is negative as in the first example above, then in the previous column the constant term in +%&' must have been positive in order for the final column numbers to add to make 0. If the constant term in $%&' were positive as in the second example above, then in the previous column the constant term in +%&' must have been negative in order for the final column numbers to add to make 0. So the constant terms in $%&' and +%&' must have opposite signs. This argument has depended on the facts that ;:* and that $%;'*3 But +%&' and $%&' both start with the same positive coefficient. Next, reading from left to right, we claim +%&' cannot change signs until $%&' changes signs. Whenever +%&' changes signs from one column to the next, $%&' must also change signs between those same two columns. But as in the second example above (columns 2 & 3 and columns 3 & 4), +%&' can keep the same sign even when $%&' does change sign. But +%&' can never change signs unless $%&' changes signs first. Now suppose in counting sign changes that at some point $%&' changes signs when +%&' does not as in the first two examples above. Then $%&' has one more sign variation, and from that point forward, $%&' will continue to lead in the sign variation count because each further time +%&' changes signs, so does $%&'3 We can rest our case in this case. The other case that needs to be considered is when counting sign changes, if we reach the end of +%&', and if at that point $%&' and +%&' have the same sign variation count, as in the third example above. Then $%&' and +%&' will have the same sign in the next to the last column, but then $%&' will change signs one more time in its last column. No matter how you look at it, $%&' has at least one more sign variation count than +%&'3 If the leading coefficient of $%&' is not positive, then factor out = 1 from $%&' and then apply the above argument to the resulting polynomial. We have already proved $%&' has at least one more sign variation than the quotient polynomial +%&'. To prove that the difference is always an odd number, we reiterate that the constant terms in both polynomials always differ in sign while the first terms always agree in sign. So when more than one sign variation occurs, it occurs at some interior coefficient. But changing the sign of any interior coefficient either raises or lowers the sign variation count by 2 because such a change applies to the term before it and to the term after it. So when the difference in the number of sign variations is more than 1, it must be an odd difference. page 15
17 18. Descartes's Rule of Signs Lemma 2. If $%&' has real coefficients, the number of positive zeros of $%&' is not greater than the number of variations in sign of the coefficients of $%&'3 Descartes's Rule of Signs Lemma 2 Proof: 1 9 B 6 denote all the positive roots of the equation $%&'*3 Then we may write $%&'  %& =, 1'%& =, 9'%& =, B'<%& =, 6'. T%&'3 Now consider the following regrouping of these factors: $%&'  %& =, 1' '%& =, 9'%& =, B'<%& =, 6'. T%&' ( 3 By Lemma 1 we know $%&' has at least one more sign variation than the rightmost factor. Let + 1%&'  '%& =, 9'%& =, B'<%& =, 6'. T%&' ( 3 Then $%&' has at least one more sign variation than +%&' 1 has. Moreover, since we may write + 1%&'  %& =, 9' '%& =, B'<%& =, 6'. T%&' ( we can again apply Lemma 1 to conclude that +%&' 1 has one at least more sign variation than the polynomial D%& =, '<%& =, '. T%&'E. Let + %&'  D%& =, '<%& =, '. T%&'E3 B 6 9 B 6 Now $%&' has at least one more sign variation than + 1%&' and + 1%&' has at least one more sign variation than + so $%&' has at least two more sign variations than + %&'3 9 9 Clearly we may continue to regroup the rightmost factors and reduce the number of factors. + 9%&'  D%& =, B'<%& =, 6'. T%&'E  %& =, B'. D%& =, I'<%& =, 6'. T%&'E3 So + %&' has one or more sign variations than + %&'  D%& =, '<%& =, '. T%&'E3 9 B I 6 We argue that for each factor we drop, $%&' has yet at least another sign variation more than the resulting rightmost factor. After dropping all 6 factors we conclude that $%&' has 6 or more sign variations than does T%&', since after dropping 6 factors, T%&' is all that remains. Now assume T%&' has 4 sign variations in its coefficients and assume $%&' has 2 sign variations in its coefficients. Then because 40*, 4/6N2implies 6N2. The number of positive zeros of $%&' is less than or equal to the number of sign variations in the coefficients of $%&'3 page 16
18 19. Descartes's Rule of Signs Lemma 3. denote 6 positive numbers and let $%&',%&=,'3 1 9 B 6 C C1 Then the coefficients of $%&' are all alternating in sign and this polynomial has exactly 6 sign variations in its coefficients. Descartes's Rule of Signs Lemma 3 Proof: Either use induction on 6 or else apply Lemma 2 to conclude that T%&'  1has 6 fewer variations in sign than $%&'3 But since 1 has no variations in sign we know $%&' must have 6 variations in sign which means all the coefficients alternate in sign. 9 9 As a simple example: %& =, 1'%& =, 9'  & = %, 1 /, 9'& / % =1 & %, 1., 9& 3 Using induction, if 61, then we note %&=,' 1 has exactly 1sign variation. Next, assume the theorem is true for any polynomial with 6 factors or less, and let 6/1 6 $%&',%&=,'%&=, C 6/1'.!,%&=,' C " C1 C1 If we consider the second factor to be the quotient polynomial then we can apply the induction assumption to conclude this quotient has exactly 6 sign variations. Next we apply Lemma 1 to $%&' to conclude that $%&' has at least one more sign variation than the quotient. This means UP $%&' has exactly 6/1 sign variations. Being a %6/1' degree polynomial, $%&' cannot have more than 6/1sign variations because it has only 6/9coefficients Descartes's Rule of Signs Lemma 4. The number of variations in sign of a polynomial with real coefficients is even if the first and last coefficients have the same sign, and is odd if the first and last coefficients have opposite signs. Descartes's Rule of Signs Lemma 4 Proof: Before giving the proof we look at one example. I B 9 $%&'  & = K& / 11& = 19& / 1J3 In this case, the first and last coefficients have the same sign and we can see that $%&' has an even number of sign changes in its coefficients; it has 4 sign changes. The degree of $%&' is I, it has J coefficients, and thus it has an a priori possibility of having at most 4 sign variations. If we were to change the sign of either the first or the last coefficient, we would have one less sign change, or an odd number of sign changes. If we were to increase the degree of $%&' by adding just one term, then we would not add a sign change unless the sign of that new term differed from the existing leading term's sign. We prove this theorem by strong induction on the degree 2 of the polynomial $%&'3 When 21, we assume $%&';&/H. If ; and Hhave the same sign then we have 0 or an even number of sign changes. If ; and H have opposite signs then we have 1 or an odd number of sign changes. So the theorem is true when 213 proof continued on the next page page 17
19 Next, assume the theorem is true whenever 2N6, and let $%&' be a polynomial of degree 6 / 13 Must show the theorem is true for $%&'3 Consider the polynomial of degree 6, obtained by dropping the leading term from $%&'3 Call this polynomial +%&'3 The theorem is assumed true for +%&' since its degree can be assumed to be either 6, or even better, less than 6. There are two cases. Case 1: +%&' 's leading and trailing terms have the same sign. Then we know by the induction assumption that +%&' has an even number of sign changes. There are only two possibilities for the sign of the leading term that was dropped. If the dropped leading term has the same sign as the leading term in +%&', then there is no sign change when this term is added back. So $%&' would still have an even number of sign changes and the leading and trailing terms of $%&' would still agree in sign. If the dropped leading term has a different sign from the leading term in then there is one additional sign change that gets added when this term is put back. So in this case $%&' would have an odd number of sign changes. But also in this case, the leading and trailing terms of $%&' would have opposite signs. Case 2: +%&' 's leading and trailing terms have opposite signs. Then we know by the induction assumption that +%&' has an odd number of sign changes. There are only two possibilities for the sign of the leading term that was dropped. If the dropped leading term has the same sign as the leading term in +%&', then there is no sign change when this term is added back. So $%&' would still have an odd number of sign changes and the leading and trailing terms of $%&' would still have opposite signs. If the dropped leading term has a different sign from the leading term in then there is one additional sign change that gets added when this term is put back. So in this case $%&' would have an even number of sign changes. But also in this case, the leading and trailing terms of $%&' would have the same signs. In either case, the theorem is true for $%&' with degree 6 / 13 This completes the proof by induction on 2. page 18
20 21. Descartes's Rule of Signs Lemma 5. If the number of positive zeros of $%&' with real coefficients is less than the number of sign variations in $%&', it is less by an even number. Descartes's Rule of Signs Lemma 5 Proof: If the leading coefficient of $%&' isn't 1, we can factor it out and just assume $%&' '%& =, 1'<%& =, 6' (. '%& = 21'<%& = 2 O' (. '%& / H 1& / 51'<%& / H V& / 5 V' ( where the, C denote all the positive zeros of $%&', the 2C denote the all the negative zeros of $%&', and the remaining factors are quadratics corresponding to all the complexconjugate paired complex zeros of $%&'. Let U be the number of sign changes in the coefficients of $%&'3 We assume 6GUand we must show there exists an even integer Wsuch that W:* and 6/WU3 9 9 Let X%&' ''%&=21'<%&=2' O (.'%& /H 1&/51'<%& /H&/5' V V (( 3 By Lemma 2, we know the X%&' polynomial has at least 6 fewer sign variations than $%&'3 We let Pbe the number of sign changes in the X%&' polynomial. Then we know P0U=6:*3 Next we note a special property of each of the irreducible quadratic factors. The discriminant 9 9 of each quadratic must be negative, so we know H C =I.1.5C G*3 So *NHC GI5C and we conclude that all the 5 C coefficients are strictly positive. Also, each %& = 2 C' factor of X%&' may be written as %& / $ C' where $ C  =2C is positive. So 9 9 we may write X%&' ''%&/$ 1'<%&/$' O (.'%& /H 1&/51'<%& /H&/5' V V (( 3 Now it is clear that the leading coefficient of X%&' is /1, and the constant term of X%&' is also positive since it is the product of all positive numbers. O The constant term of X%&'  3, $ C4!, 5C ". By Lemma 4, the number of sign variations C1 C1 V in the coefficients of X%&' is even. Pis even. From above we have P0U=6:*3 Therefore P/60U3 Now if it happens that P/6Uthen we let WPand we are done. Otherwise, if P/6:Uthen we have to argue about the first 6 factors in $%&'3 Having just established that the constant term in X%&' is positive, the sign of the constant term in $%&' is 6 6 the sign of % &!, 6 =1., 6"  the sign of % =1 & since all the, C values are positive. If 6 is even C1 then by Lemma I$%&', has an even number of sign variations in its coefficients which means U is even. If 6 is odd then again by Lemma 4 we conclude U is odd. So 6 and Uare even together or are odd together. Now consider that P/6 =U : *3 What kind of a positive number is this? Well Pis even, and if 6and Uare both even then P/6=Umust be even. By the same token, if 6 and Uare both odd, then 6 =U is still even, and since P is always even, P/%6=U' must be even. Therefore P/6=U9Yfor some positive integer Y3 Then %P = 9Y' / 6  U3 Let W  %P = 9Y'3 All that remains is to show that %P = 9Y' is a positive even integer. First, %P=9Y'U=6and since U=6:*, we know %P=9Y' is a positive integer. Second, both P and 9Y are even, so their difference %P = 9Y' is even. In any case, there exists a positive even integer W such that W/6  U3 So when U is larger than 6, it is larger by a positive even integer. page 19
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