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) = a_nxⁿ + a_n−1xⁿ⁻¹ + … + a₂x² + a₁x + a₀

Polynomials can be classified as monomials, binomials, and trinomials based on the number of terms present. For example, 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 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 a_nxⁿ + a_n-1x^n-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:

[Tex]x_1=x_0-\frac{f(x_0)}{f'({x_0})} [/Tex]

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 of Newton’s Method

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'(x_n). The x- intercept occurs where y = 0.

Thus, by setting y = 0, we have:

[Tex]x_1=x_0-\frac{f(x_0)}{f'({x_0})} [/Tex]

Hence proved.

Sample Problems on Newton’s Method for Finding Roots

Question 1. Starting with x₀ = 5, find the next root of the equation x³ − 7x² + 8x − 3 = 0.

Solution:

As per Newton’s method, [Tex]x_1=x_0-\frac{f(x_0)}{f'({x_0})}  [/Tex].

Given: x₀ = 5, f(x) = x³ − 7x² + 8x − 3 = 0

⇒ f'(x) = 3x² − 14x + 8

Now, x₁ = [Tex]x_0-\frac{f(5)}{f'({5})} [/Tex]

= 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, [Tex]x_1=x_0-\frac{f(x_0)}{f'({x_0})}. [/Tex]

Given: x₀ = 5, f(x) = x³ − x² − 15x + 1 = 0

⇒ f'(x) = 3x² − 2x − 15

Now, x₁ = [Tex]x_0-\frac{f(-3.5)}{f'({-3.5})} [/Tex]

= −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, [Tex]x_1=x_0-\frac{f(x_0)}{f'({x_0})}  [/Tex].

Given: x0 = 5, f(x) = x² − 2 = 0

⇒ f'(x) = 2x

Now, x₁ = [Tex]x_0-\frac{f(2)}{f'({2})} [/Tex]

= 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, [Tex]x_1=x_0-\frac{f(x_0)}{f'({x_0})}. [/Tex]

Given: x0 = 5, f(x) = −x³ + 4x² − 2x + 2 = 0.

Now, x1 = [Tex]x_0-\frac{f(2)}{f'({2})} [/Tex]

= 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, [Tex]x_1=x_0-\frac{f(x_0)}{f'({x_0})}. [/Tex]

Given: x0 = 5, f(x) = −x3 + 4×2 − 2x + 2 = 0.

Now, x1 = [Tex]x_0-\frac{f(2)}{f'({2})} [/Tex]

= 1 − 3/3

⇒ x1 = 0

Read More:

• Newton Raphson Method | Definition, Formula, Examples
• Root Finding Algorithm
• Program for Newton Raphson Method

Practice Problems on Newton’s Method for Finding Roots

1. Starting with x₀ = 0.5. Find the square roots by Newton’s Method for Finding Roots of equation cos(x) – x

2. Starting with x₀ = 1. Find the square roots by Newton’s Method for Finding Roots of equation e^x – 3x

3. Starting with x₀ = 0. Find the square roots by Newton’s Method for Finding Roots of equation x³ – 2x + 2

4. Starting with x₀ = 2. Find the square roots by Newton’s Method for Finding Roots of equation ln(x) – 1

FAQs on Newton’s Method for Finding Roots

What is Newton’s Method?

Newton’s Method, also known as the Newton-Raphson method, is an iterative numerical technique used to find approximate solutions (roots) of a real-valued function. It is based on the idea of linear approximation and uses the function’s derivative to find successive approximations to the root.

How does Newton’s Method work?

Newton’s Method starts with an initial guess x0x_0x0​ for the root of the function f(x). The next approximation is given by the formula:

[Tex]x_1=x_0-\frac{f(x_0)}{f'({x_0})} [/Tex]

What are the conditions for the convergence of Newton’s Method?

• The function f(x) must be differentiable in the interval containing the root.
• The derivative f′(x) should not be zero at the root.
• The initial guess x₀​ should be sufficiently close to the actual root.
• The function should not have inflection points or multiple roots near the initial guess.