structural:expansion
Example

Consider the polynomial p 6 (x) = (x + 8) (x + 5) (x + 3) (x − 2) (x − 3) (x − 7) {\displaystyle p_{6}(x)=(x+8)(x+5)(x+3)(x-2)(x-3)(x-7)} which can be expanded to p 6 (x) = x 6 + 4 x 5 − 72 x 4 − 214 x 3 + 1127 x 2 + 1602 x − 5040. {\displaystyle p_{6}(x)=x^{6}+4x^{5}-72x^{4}-214x^{3}+1127x^{2}+1602x-5040.}

From the above we know that the largest root of this polynomial is 7 so we are able to make an initial guess of 8. Using Newton's method the first zero of 7 is found as shown in black in the figure to the right. Next p (x) {\displaystyle p(x)} is divided by (x − 7) {\displaystyle (x-7)} to obtain p 5 (x) = x 5 + 11 x 4 + 5 x 3 − 179 x 2 − 126 x + 720 {\displaystyle p_{5}(x)=x^{5}+11x^{4}+5x^{3}-179x^{2}-126x+720} which is drawn in red in the figure to the right. Newton's method is used to find the largest zero of this polynomial with an initial guess of 7. The largest zero of this polynomial which corresponds to the second largest zero of the original polynomial is found at 3 and is circled in red. The degree 5 polynomial is now divided by (x − 3) {\displaystyle (x-3)} to obtain p 4 (x) = x 4 + 14 x 3 + 47 x 2 − 38 x − 240 {\displaystyle p_{4}(x)=x^{4}+14x^{3}+47x^{2}-38x-240} which is shown in yellow. The zero for this polynomial is found at 2 again using Newton's method and is circled in yellow. Horner's method is now used to obtain p 3 (x) = x 3 + 16 x 2 + 79 x + 120 {\displaystyle p_{3}(x)=x^{3}+16x^{2}+79x+120} which is shown in green and found to have a zero at − 3. This polynomial is further reduced to p 2 (x) = x 2 + 13 x + 40 {\displaystyle p_{2}(x)=x^{2}+13x+40} which is shown in blue and yields a zero of − 5. The final root of the original polynomial may be found by either using the final zero as an initial guess for Newton's method, or by reducing p 2 (x) {\displaystyle p_{2}(x)} and solving the linear equation. As can be seen, the expected roots of − 8, − 5, − 3, 2, 3, and 7 were found.