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In line with the corevalue to be ‘First’, the company intends to expand its market position. com 51 Click on the ad to read more Methods for finding (Real or Complex) Zeros in PolynomialsPosition of roots of polynomials in the complex plane Figure 9: The graph of P (x) = x19 − 4x13 + 6x10 + 20x5 − 2x3 − 2x2 − 4. Furthermore, ( ) 1 1 3 1 1 19 = y 19 + y 17 − 5y 14 − y 9 + y 6 − , − y P 4 y 2 2 4 so { } 1 3 1 −N = min 0, 1, , −5, − , 1, − = −5, 2 2 4 thus N = 5, and { } 1 3 1 −M = min 0, , 5, , −1, = −1, thus M = 1, 2 2 4 ] [ [ ] 1 1 ∪ , +∞ .

Pr ≥ 1, , where α1 , . . , αr denote the r mutually diﬀerent roots. Then P ′ (z) (21) = = p1 −1 a · p1 (z − α1 ) p p2 (z − α2 ) pr · · · (z − αr ) p −1 p +a · p2 (z − α1 ) 1 (z − α2 ) 2 · · · (z − αr ) r +··· p p p −1 +a · pr (z − α1 ) 1 · · · (z − αr−1 ) r−1 · · · (z − αr ) r p1 −1 a (z − α1 ) pr −1 · · · (z − αr ) {p1 (z − α2 ) · · · (z − αr ) + · · · + pr (z − α1 ) · · · (z − αr−1 )} . com 38 Methods for finding (Real or Complex) Zeros in Polynomials Some solution formulæ of roots of polynomials If we put z = αj into the latter factor of (21), all terms disappear with the exception of (22) pj (αj − α1 ) · · · (αj − αj−1 ) · (αj − αj+1 ) · · · (αj − αr ) ̸= 0.

Furthermore, Hm (x) = 1 > 0 for all x. Hence, W (x) is constant in each of the intervals mentioned above. 1 Let x1 , . . , xN be given as above. Fore every i = 1, . . , N , the value of W (xi ) is equal to the value of W (x) in the adjacent interval to the right of xi . In the adjacent interval to the left of xi the value of W (x) is given by if xi is not a root of P (x), W (xi ) , W (x) = if xi is a root of P (x). com 63 Methods for finding (Real or Complex) Zeros in PolynomialsPosition of roots of polynomials in the complex plane Main case H(x) H1 (x) ..