5 NEW SIMULTANEOUS METHODS FOR FINDING ALL ZEROS OF A POLYNOMIAL From (8), we obtain P ( z$_{i}$ ) P $^{′}$( z$_{i}$ ) ( P $^{′′}$( z$_{i}$ ) P ( z$_{i}$ ) 2 P $^{′}$( z$_{i}$ ) 2 O$_{M}$ ( e ) + O$_{M}$ ( e $^{2}$) ) = 1 2 ( e$_{i}$ + P $^{′′}$( r$_{i}$ ) 2 P $^{′}$( r$_{i}$ ) e 2 i + O$_{M}$ ( e $^{3}$) ) $^{2}$( 1 − P $^{′′}$( r$_{i}$ ) P $^{′}$( r$_{i}$ ) e$_{i}$ + P $^{′′′}$( r$_{i}$ ) 2 P $^{′}$( r$_{i}$ ) e 2 i + O$_{M}$ ( e $^{3}$) ) 3 × P $^{′′}$( r$_{i}$ ) P $^{′}$( r$_{i}$ ) + P $^{′′′}$( r$_{i}$ ) P $^{′}$( r$_{i}$ ) e$_{i}$ + P $^{′′′′}$( r$_{i}$ ) 2 P $^{′}$( r$_{i}$ ) e 2 i + O$_{M}$ ( e $^{3}$) ) O$_{M}$ ( e ) + O$_{M}$ ( e $^{3}$) = 1 2 ( O$_{M}$ ( e )) $^{2}$(1 + O$_{M}$ ( e )) 3 ( P $^{′′}$( r$_{i}$ ) P $^{′}$( r$_{i}$ ) + O$_{M}$ ( e ) ) O$_{M}$ ( e ) + O$_{M}$ ( e $^{3}$) = O$_{M}$ ( e $^{3}$) . □ From Lemma 1, we have the following theorem. Theorem 1. If the approximate zero x$_{i}$ grounded from Method 1 is close enough to r$_{i}$ and the order of e$_{i}$ is the same, then the order of ˆ e$_{i}$ is identical, and | ˆ e$_{i}$ | = O$_{M}$ ( | e | $^{3}$) is formed. Proof. We easily see that the following equation is satisfied. That is, If Q ( z ) = P ( z ) − ∏ n j $_{=1}$( z − z$_{j}$ ), then Q ( z ) is a polynomial of order n − 1, and Q ( z$_{i}$ ) = P ( z$_{i}$ ) for all i . Therefore, Q ( z ) is the Lagrange interpolation of points z$_{1}$, z$_{2}$, . . . , z$_{n}$ , and so we have P ( z$_{i}$ ) = n ∏ j =1 ( z$_{i}$ − r$_{j}$ ) = ( z$_{i}$ − r$_{i}$ ) ∏ j = i ( z$_{i}$ − r$_{j}$ ) = e$_{i}$ ∏ j = i ( z$_{i}$ − r$_{j}$ ) = O$_{M}$ ( e ) . W$_{i}$ = W$_{i}$ ( z$_{i}$ ) = O$_{M}$ ( P ( z$_{i}$ )) = O$_{M}$ ( e ) . (9) Q ( z ) = n ∑ j $_{=1}$( P ( z$_{j}$ ) ∏ k = j z − z$_{k}$ z$_{j}$ − z$_{k}$ ) = n ∑ j $_{=1}$(( W$_{j}$ ∏ k = j ( z$_{j}$ − z$_{k}$ ) ) ∏ k = j z − z$_{k}$ z$_{j}$ − z$_{k}$ ) = n ∑ j $_{=1}$( W$_{j}$ ∏ k = j ( z − z$_{k}$ ) ) = ( n ∑ j =1 W$_{j}$ z − z$_{j}$ ) n ∏ k =1 ( z − z$_{k}$ ) .