One task performed by scientists and engineers is finding the roots of a polynomial f(x), i.e. given
f(x) = cnxn + cn–1 xn–1 + cn–2 xn–2 + . . . + c2x2 + c1x1 + c0x0 = 0
find x0 such that f(x0) = 0. Recall that a polynomial f(x) of degree n has a maximum n distinct roots. Some roots may be real numbers and some may be complex numbers.
Factoring is one way of determining the roots of a polynomial f(x), if the polynomial is factorable.
| Example: |
Consider
f(x) = 2x4 – 7x3 + 4x2 + 7x – 6 = 0 f(x) = 2x4 + 4x2 – 6 – 7x3 + 7x = 0 f(x) = 2 ( x4 + 2x2 – 3 ) – 7x ( x2 – 1 ) = 0 f(x) = 2 ( x2 + 3 ) ( x2 – 1 ) – 7x ( x2 – 1 ) = 0 f(x) = [ 2 ( x2 + 3 ) – 7x ] ( x2 – 1 ) = 0 f(x) = ( 2x2 + 6 – 7x ) ( x2 – 1 ) = 0 f(x) = ( 2x2 – 7x + 6 ) ( x2 – 1 ) = 0 f(x) = ( 2x – 3 ) ( x – 2 ) ( x + 1 ) ( x – 1 ) = 0 x1 = 3/2 x2 = 2 x3 = – 1 x4 = 1 |
However, not all polynomials are factorable.
Example: f(x) = x5 – 2x3 – 5x2 + 2 = 0 is not factorable.
If a non–factorable polynomial is a quadratic function, then the quadratic formula is used. For polynomials of higher degrees, other more general techniques are required. We shall look at several techniques for estimating roots of polynomials, namely the Bisection Method, Regula–Falsi Method, and Newton's Method. To simplify the problem, find the first positive root of a given polynomial f(x). For the initial interval [a, b], find the unit interval containing the first positive root.
However, let us first take a look at an efficient way of evaluating a polynomial using Horner's Method.