Properties ad Tests of Zeros of Polyomial Fuctios The Remaider ad Factor Theorems: Sythetic divisio ca be used to fid the values of polyomials i a sometimes easier way tha substitutio. This is show by the ext theorem. If the polyomial P(x) is divided by x c, the the remaider is the value P(c). We leared that if c is a zero of P, tha x c is a factor of P(x). The ext theorem restates this fact i a more useful way. Factor Theorem: c is a zero of P if ad oly if x c is a factor of P(x). A importat cosequece of the Factor Theorem is that fidig the zeros of a polyomial is really the same thig as factorig it ito liear factors. I this sectio we will study more methods that help us fid the real zeros of a polyomial, ad thereby factor the polyomial. Ratioal Zeros of Polyomials: The ext theorem gives a method to determie all possible cadidates for ratioal zeros of a polyomial fuctio with iteger coefficiets. Ratioal Zeros Theorem: If the polyomial ( ) P x = ax + a x +... + ax+ a0 has iteger coefficiets, the every ratioal zero of P is of the form p q where p is a factor of the costat coefficiet a0 ad q is a factor of the leadig coefficiet a Example : List all possible ratioal zeros give by the Ratioal Zeros Theorem of P(x) = 6x 4 + 7x 3-4 (but do t check to see which actually are zeros). Solutio: Step : First we fid all possible values of p, which are all the factors of a 0 = 4. Thus, p ca be ±, ±2, or ±4.
Example (Cotiued): Step 2: Next we fid all possible values of q, which are all the factors of a = 6. Thus, q ca be ±, ±2, ±3, or ±6. Step 3: Now we fid the possible values of p q by makig combiatios of the values we foud i Step ad Step 2. Thus, p q the form factors of 4 factors of 6. The possible p q are will be of 2 4 2 4 2 4 2 ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ± 4 2 2 2 3 3 3 6 6 6 Step 4: Fially, by simplifyig the fractios ad elimiatig duplicates, we get the followig list of possible values for p q. 2 4 ±, ± 2, ± 4, ±, ±, ±, ±, ± 2 3 3 3 6 Now that we kow how to fid all possible ratioal zeros of a polyomial, we wat to determie which cadidates are actually zeros, ad the factor the polyomial. To do this we will follow the steps listed below. Fidig the Ratioal Zeros of a Polyomial:. Possible Zeros: List all possible ratioal zeros usig the Ratioal Zeros Theorem. 2. Divide: Use Sythetic divisio to evaluate the polyomial at each of the cadidates for ratioal zeros that you foud i Step. Whe the remaider is 0, ote the quotiet you have obtaied. 3. Repeat: Repeat Steps ad 2 for the quotiet. Stop whe you reach a quotiet that is quadratic or factors easily, ad use the quadratic formula or factor to fid the remaiig zeros.
Sythetic Divisio: Sythetic divisio is a shortcut method of performig log divisio that ca be used whe the divisor is a first degree polyomial of the form x c. I sythetic divisio we write oly the essetial part of the log divisio table. To illustrate, compare these log divisio ad sythetic divisio tables, i which we divide 3x 3 4x + 2 by x : Note that i sythetic divisio we abbreviate 3x 3 4x + 2 by writig oly the coefficiets: 3 0 4 2, ad istead of x, we simply write. (Writig istead of allows us to add istead of subtract, but this chages the sig of all the umbers that appear i the yellow boxes.) To divide a x + a - x - +... + a x + a 0 by x c, we proceed as follows: Here b - = a, ad each umber i the bottom row is obtaied by addig the umbers above it. The remaider is r ad the quotiet is b x + b x +... + bx+ b 2 2 0
Example 2: Fid the quotiet ad the remaider of x 7x 6x usig sythetic divisio. x + 2 4 2 Solutio: Step : We put x + 2 i the form x c by writig it as x ( 2). Use this ad the coefficiets of the polyomial to obtai 2 0 7 6 0 Note that we used 0 as the coefficiet of ay missig powers. Step 2: Next, brig dow the. 2 0 7 6 0 Step 3: Now, multiply 2 by to get 2, ad add it to the 0 i the first row. The result is 2. 2 0 7 6 0 2 2 Step 4: Next, 2( 2) = 4. Add this to the 7 i the first row. 2 0 7 6 0 2 4 2 3
Example 2 (Cotiued): Step 5: 2( 3) = 6. Add this to the 6 i the first row. 2 0 7 6 0 2 4 6 2 3 0 Step 6: Fially, 2(0) = 0, which is added to 0 to get 0. 2 0 7 6 0 2 4 6 0 2 3 0 0 The coefficiets of the quotiet polyomial ad the remaider are read directly from the bottom row. Also, the degree of the quotiet will always be oe less tha the degree of the divided. Thus, Q(x) = x 3 2x 2 3x ad R(x) = 0. Example 3: Fid all real zeros of the polyomial P(x) = 2x 4 + x 3 6x 2 7x 2. Solutio: Step : First list all possible ratioal zeros usig the Ratioal Zeros Theorem. For the ratioal umber p q to be a zero, p must be a factor of a 0 = 2 ad q must be a factor of a = 2. Thus the possible ratioal zeros, p q, are ±, ± 2, ± 2 Step 2: Now we will use sythetic divisio to evaluate the polyomial at each of the cadidates for ratioal zeros we foud i Step. Whe we get a remaider of zero, we have foud a zero.
Example 3 (Cotiued): 2 6 7 2 2 3 3 0 2 3 3 0 2 2 6 7 2 2 5 2 2 5 2 0 Sice the remaider is ot zero, + is ot a zero Sice the remaider is zero, is a zero This also tells us that P factors as 2x 4 + x 3 6x 2 7x 2 = (x + )(2x 3 x 2 5x 2) Step 3: We ow repeat the process o the quotiet polyomial 2x 3 x 2 5x 2. Agai usig the Ratioal Zeros Theorem, the possible ratioal zeros of this polyomial are ±, ± 2, ±. 2 Sice we determied that + was ot a ratioal zero i Step 2, we do ot eed to test it agai, but we should test agai. 2 5 2 2 3 2 2 3 2 0 Thus, P factors as Sice the remaider is zero, is agai a zero 2x 4 + x 3 6x 2 7x 2 = (x + )(2x 3 x 2 5x 2) = (x + ) (x + )(2x 2 3x 2) = (x + ) 2 (2x 2 3x 2)
Example 3 (Cotiued): Step 4: At this poit the quotiet polyomial, 2x 2 3x 2, is quadratic. This factors easily ito (x 2)(2x + ), which tells us we have zeros at x = 2 ad x =, ad that P factors as 2 2x 4 + x 3 6x 2 7x 2 = (x + )(2x 3 x 2 5x 2) = (x + ) (x + )(2x 2 3x 2) = (x + ) 2 (2x 2 3x 2) = (x + ) 2 (x 2)(2x + ) Step 5: Thus the zeros of P(x) = 2x 4 + x 3 6x 2 7x 2 are x =, x = 2, ad x =. 2 Fudametal Theorem of Algebra: I 799 the Germa mathematicia C. F. Gauss proved the Fudametal Theorem of Algebra. This Theorem forms the basis for much of our work i factorig polyomials ad solvig polyomial equatios. Fudametal Theorem of Algebra: Every polyomial P( x) = a x + a x +... + a x+ a (, a 0) 0 with complex coefficiets has at least oe complex zero. Because ay real umber is also a complex umber, the theorem applies to polyomials with real coefficiets as well. The coclusio of the Fudametal Theorem of Algebra is that for every polyomial P(x), there is a complex umber c such that P( c ) = 0. From the Factor Theorem (Sectio 5.2), this tells us x c is a factor of P(x). Thus we ca write ( ) = ( ) ( ) P x x c Q x where Q(x) has degree. If the quotiet Q(x) has degree we ca repeat the procedure of obtaiig a factor ad a quotiet with degree less tha the previous quotiet util we arrive at the complete factorizatio of P(x). This process is summarized by the ext theorem.
Complete Factorizatio Theorem: If P(x) is a polyomial of degree > 0, the there exist complex umbers a, c, c 2,... c (with a 0) such that ( ) = ( )( ) ( ) P x a x c x c x c 2... I the Complete Factorizatio Theorem the umbers c, c 2,... c are the zeros of P. These zeros eed ot all be differet. If the factor x c appears k times i the complete factorizatio of P(x), the we say that c is a zero of multiplicity k. The ext theorem follows from the Complete Factorizatio Theorem. Zeros Theorem: Every polyomial of degree has exactly zeros, provided that a zero of multiplicity k is couted k times. Descartes Rule of Sigs: I may cases, we will have a legthy list of possible ratioal zeros of a polyomial. A theorem that is helpful i elimiatig cadidates is Descartes Rule of Sigs. I the theorem, variatio i sig is a chage from positive to egative, or egative to positive i successive terms of the polyomial. Missig terms (those with 0 coefficiets) are couted as o chage i sig ad ca be igored. For example, has two variatios i sig. Descartes Rule of Sigs: Let P be a polyomial with real coefficiets. The umber of positive real zeros of P(x) is either equal to the umber of variatios i sig i P(x) or is less tha that by a eve whole umber. 2. The umber of egative real zeros of P(x) is either equal to the umber of variatios i sig i P( x) or is less tha that by a eve whole umber.
Example 4: Use Descartes Rule of Sigs to determie how may positive ad how may egative real zeros P(x) = 6x 3 + 7x 2 3x 2 ca have. The determie the possible total umber of real zeros. Solutio: Step : First we will cout the umber of variatios i sig of ( ) 3 2 P x = 6x + 7x 3x 2. Sice there is oly oe variatio, P(x) has oe positive real zero. Step 2: Now we will cout the umber of variatios i sig of ( ) 3 2 P x = 6x + 7x + 3x 2. P( x) has two variatios i sig, thus P(x) has two or zero egative real zeros. Step 3: Fially by combiig our fidigs i the previous steps, P(x) has either oe or three real zeros.