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Math Lesson 11.1.7 - Finding the Zeroes of Polynomials through Iterative Methods

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Welcome to our Math lesson on Finding the Zeroes of Polynomials through Iterative Methods, this is the seventh lesson of our suite of math lessons covering the topic of The Definition of Monomials and Polynomials, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Finding the Zeroes of Polynomials through Iterative Methods

We can find the zeroes of higher-order polynomials by applying any of the iterative methods explained in tutorial 9.4. For example, we can use intuition to identify two values of the variable that give numbers with opposite signs as results. Since the given zero of the polynomial lies in that interval, we can narrow it until you obtain the zero of the polynomial to the desired precision. Let's consider an example to clarify this point.

Example 7

Identify one zero of following polynomial and write it at one decimal place.

P(x) = x3 - 3x2 + 2x - 7

Solution 7

Finding the zeroes of a polynomial means solving the equation P(x) = 0. Hence, we have to solve the cubic equation

x3 - 3x2 + 2x - 7 = 0

where the maximum number of zeroes (roots) is 3.

We can begin with a wide range of values, for example, 0 and 5. Thus, for x = 0, we have

P(0) = 03-3 ∙ 02 + 2 ∙ 0 - 7
= -7

and

P(5) = 53-3 ∙ 52 + 2 ∙ 5 - 7
= 125 - 3 ∙ 25 + 10 - 7
= 125 - 75 + 10 - 7
= 53

Since the two values obtained have opposite signs, it is certain that we have a zero for the given polynomial in this range. Thus, combining the change of sign method and the half-interval method, yields

P(2.5) = 2.53-3 ∙ 2.52 + 2 ∙ 2.5 - 7
= 15.625 - 3 ∙ 6.25 + 2 ∙ 2.5 - 7
= 15.625 - 18.75 + 5 - 7
= -5.125

Hence, the zero of the polynomial is between x = 2.5 and x = 5. Yet, for x = 3.75, we have

P(3.75) = 3.753-3 ∙ 3.752 + 2 ∙ 3.75 - 7
= 52.73 - 3 ∙ 14.06 + 2 ∙ 3.75 - 7
= 52.73 - 42.18 + 7.50 - 7
= 11.05

For x = 3.1 (it is not necessary to use the exact half interval as this gives numbers with many decimal places but we can use values that are close to half of the given interval), we obtain

P(3.1) = 3.13 - 3 ∙ 3.12 + 2 ∙ 3.1 - 7
= 29.8 - 3 ∙ 9.6 + 2 ∙ 3.1 - 7
= 29.8 - 28.8 + 6.2 - 7
= -0.8

Again, for x = 3.4, we obtain

P(3.4) = 3.43-3 ∙ 3.42 + 2 ∙ 3.4 - 7
= 39.3 - 3 ∙ 11.6 + 2 ∙ 3.4 - 7
= 39.3 - 34.8 + 6.8 - 7
= 4.3

Now that we know that the zero of the give polynomial is between 3.1 and 3.4, we check for x = 3.25. Thus,

P(3.25) = 3.253 - 3 ∙ 3.252 + 2 ∙ 3.25 - 7
= 34.3 - 3 ∙ 10.6 + 2 ∙ 3.25 - 7
= 34.3 - 31.8 + 6.5 - 7
= -3

Hence, the approximate zero of the given polynomial written to one decimal place is x = 3.3, as it is the only option for numbers with one decimal place between 3.25 and 3.4.

More The Definition of Monomials and Polynomials Lessons and Learning Resources

Polynomials Learning Material
Tutorial IDMath Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
11.1The Definition of Monomials and Polynomials
Lesson IDMath Lesson TitleLessonVideo
Lesson
11.1.1The Definition of Monomials
11.1.2The Definition of Polynomials
11.1.3The Degree of Polynomials
11.1.4The Names of Polynomials by Degree
11.1.5Finding the Value of Polynomials
11.1.6Finding the Zeroes of a Polynomial
11.1.7Finding the Zeroes of Polynomials through Iterative Methods
11.1.8The Homogenous and Non-Homogenous Polynomials

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