Polynomial is composed of two words: Poly (which means “many”) and Nominal (which means “terms.”). A polynomial is a mathematical equation made up of variables, constants, and exponents that are mixed using operations like addition, subtraction, multiplication, and division. The general form of a polynomial is
f(x) = anxn + an−1xn−1 + … + a2x2 + a1x + a0
Polynomials can be classified as monomials, binomials and trinomials based upon the number of terms present. For example, the terms like x, 13y, 39, etc. are all monomials while terms like x2 + x, x10 – x4, etc. are termed as binomials because they consist of two terms. Similarly, such polynomials as having only three terms are termed as trinomials.
Roots of a Polynomial
The roots of a polynomial are the solutions to the given polynomial for which the unknown variable’s value must be determined. We can evaluate the value of a polynomial to zero if we know the roots. A polynomial of degree ‘n’ in variable x is an equation of the type anxn + an-1xn-1 +…… + a1x + a0, where each variable has a constant as its coefficient. The phrase refers to each variable in an expression that is separated by an addition or subtraction sign. The greatest power of a polynomial variable is defined as the degree of the polynomial.
Newton’s Method of Finding Roots of a Polynomial
Newton’s method formula is used for finding the roots of a polynomial by iterating from one root to the next. Calculating the roots by this approach takes a long time for polynomials of greater degree, but for polynomials of lower degree, the results are quite rapid and near to the true roots of the equation. Using this strategy, we can identify the consecutive roots of an equation if we know any one of its roots. The formula for Newton’s method of finding the roots of a polynomial is as follows:
where,
- x0 is the initial value
- f(x0) is the function value at the initial value
- f'(x0) is the first derivative of the function value at initial value.
Derivation
Newton’s method begins with an initial guess that is relatively near to the correct root (solution), and then utilize the tangent line to acquire an even better x-intercept than our first guess or beginning point.
Assume we’re looking for the root (or x-intercept) of f(x). This indicates we’re looking for an in the image below. The trick of Newton’s method is to draw a tangent line to the graph y = f(x) at the point (b, f(b)).
The crucial calculation in each stage of Newton’s technique, as we just saw, is to discover where the tangent line to y = f(x) at the point (x1, y1) crosses the x-axis.
The point gradient form of equation of a line with slope m and passing through the point (x0, y0) is given as:
y – y0 = m(x – x0).
In this situation, the line has a slope of f'(xn). The x- intercept occurs where y = 0.
Thus, by setting y = 0, we have:
Hence proved.
Sample Problems
Question 1. Starting with x0 = 5, find the next root of the equation x3 − 7x2 + 8x − 3 = 0.
Solution:
As per Newton’s method,
. Given: x0 = 5, f(x) = x3 − 7x2 + 8x − 3 = 0
⇒ f'(x) = 3x2 − 14x + 8
Now, x1 =
= 5 − (−13)/13
= 5 + 1
⇒ x1 = 6
Question 2. Starting with x0 = −3.5, find the next root of the equation x3 − x2 − 15x + 1 = 0.
Solution:
As per Newton’s method,
Given: x0 = 5, f(x) = x3 − x2 − 15x + 1 = 0
⇒ f'(x) = 3x2 − 2x − 15
Now, x1 =
= −3.5 − (−1.625)/28.75
⇒ x1 = −3.443
Question 3. Starting with x0 = 2, find the next root of the equation x2 − 2 = 0.
Solution:
As per Newton’s method,
. Given: x0 = 5, f(x) = x2 − 2 = 0
⇒ f'(x) = 2x
Now, x1 =
= 2 − 2/4
⇒ x1 = 1.5
Question 4. Starting with x0 = 1, find the next root of the equation −x3 + 4x2 − 2x + 2 = 0.
Solution:
As per Newton’s method,
Given: x0 = 5, f(x) = −x3 + 4x2 − 2x + 2 = 0.
Now, x1 =
= 1 − 3/3
⇒ x1 = 0
Question 5. Starting with x0 = 8, find the next root of the equation x3 − 10x2 + 9x − 12 = 0.
Solution:
As per Newton’s method,
Given: x0 = 5, f(x) = −x3 + 4×2 − 2x + 2 = 0.
Now, x1 =
= 1 − 3/3
⇒ x1 = 0