Roots of Algebraic and Transcendental Equations - Numerical Structural Analysis - page 10

Table 1.8. Example 1.6 Synthetic division

− 1 1 − 6 11 − 6

0 − 1 7 − 18

− 1 1 − 7 18 − 24

0 − 1 8

1 − 8 26

Since the remainder, R , is 26, f ′( - 1) =R*1!= 26. The first derivative of the
polynomial evaluated at - 1 is 26. Performing a check as follows:

′ () =−+=
′ () = () − () +=

fx x 32 10

2

x

2

f 131 2111 26

Divide the remaining polynomial again by - 1 to find f− ′′( - 1) .

Table 1.9. Example 1.6 Synthetic division

− 1 1 − 6 11 − 6

0 7 − 1 − 18

1 18 − 1 − 7 − 24

0 − 1 8

− 1 1 − 8 26

0 − 1

1 − 9

Since the remainder, R , is - 9, f ′′( - 1)= R*2!= - 9(1)( 2) = - 18 and the
second derivative of the polynomial evaluated at - 1 is - 18. Performing a
check as follows:

′′ () =−=
′′ () = () −=−

f x x
f

6 20
16112 8

1.5 incREMEntAL SEARcH MEtHOD

The incremental search method is a simple and quick way to find the
approximate location of real roots to algebraic and transcendental