SOLVING POLYNOMIALS OF HIGHER DEGREE WITH EXAMPLES

Please study Method of Solving Polynomials if you have not already done so. There we gave introduction to Algebra Equations of higher degree and discussed the method of solving them.

We developed the relations between the roots and coefficients of the equation and gave Formulas for the same. That knowledge is a prerequisite here. So please study them before proceeding further.

Get The Best Grades With the Least Amount of Effort : Solving Polynomials

Here is a collection of proven tips, tools and techniques to turn you into a super-achiever - even if you've never thought of yourself as a "gifted" student.

The secrets will help you absorb, digest and remember large chunks of information quickly and easily so you get the best grades with the least amount of effort.

If you apply what you read from the above collection, you can achieve best grades without giving up your fun, such as TV, surfing the net, playing video games or going out with friends!

Solve the following problem on Solving Polynomials.

Solve the Algebra Equation 6x^{4} - 29x^{3} + 40x^{2} - 7x - 12 = 0, if the product of the two of its roots is equal to 2.

Solution to Example 1 Solving Polynomials: The given Algebra Equation is 6x^{4} - 29x^{3} + 40x^{2} - 7x - 12 = 0 Dividing both sides of the equation by 6, we get x^{4} - (29⁄6)x^{3} + (20⁄3)x^{2} - (7⁄6)x - 2 = 0 Comparing this with x^{4} + p_{1}x^{3} + p_{2}x^{2} + p_{3}x + p_{4} = 0, we get p_{1} = - (29⁄6); p_{2} = (20⁄3); p_{3} = -(7⁄6); p_{4} = -2

Also we have s_{1} = -p_{1} = (29⁄6);s_{2} = p_{2} = (20⁄3);s_{3} = -p_{3} = (7⁄6);s_{4} = p_{4} = -2;

By data, product of two of its roots = 2. But, s_{4} = product of all the four roots = -2 ⇒ (2)(product of the other two roots) = -2 ⇒ product of the other two roots = -1. ∴ Let the roots be α, 2⁄α, β, -1⁄β

Solution to Example 2 Solving Polynomials: The given algebra equation is x^{4} - 16x^{3} + 86x^{2} - 176x + 105 = 0

Let f(x) = x^{4} - 16x^{3} + 86x^{2} - 176x + 105 Constant term = 105 has factors 1, 3, 5, 7, -1, -3, -5, -7.Let us verify whether f(a) = 0, where a = one of those factors. f(1) = 1 -16 + 86 -176 + 105 = 192 - 192 = 0 ⇒ (x - 1) is a factor of f(x). ⇒ x = 1 is a root of f(x) = 0. You can use synthetic division here, as explained in Algebra Factoring (link given above.)

⇒ f(x) ÷ (x - 1) = x^{3} - 15x^{2} + 71x - 105 Let g(x) = x^{3} - 15x^{2} + 71x - 105 g(3) = 3^{3} - 15(3)^{2} + 71(3) -105 = 27 - 135 + 213 - 105 = 0 ⇒ (x - 3) is a factor of g(x). ⇒ x = 3 is a root of g(x) = 0. Let α and β be the other roots of g(x) = 0 s_{1} = 3 + α + β = -p_{1} = 15⇒ α + β = 12......(i)

s_{2} = (3)(α) + (3)(β) + (α)(β) = p_{2} = 71⇒ 3(α + β) + (α)(β) = 71.....(ii) Using (i) in (ii), we get 3(12) + (αβ) = 71 ⇒ αβ = 35.....(iii)

Using (i) and (iii), we can find (α - β) (α - β)^{2} = (α + β)^{2} - 4αβ= 12^{2} - 4(35) = 144 - 140 = 4 ⇒ (α - β) = ±2 ......(iv) (i) + (iv) gives 2α = 14 or 10 ⇒ α = 7 or 5. using these in (i), we get β = 12 - 7 or 12 - 5 = 5 or 7. These are same as values of α ⇒ the roots are 3, 5, 7. Let us verify whether these satisfy s_{3} or not. s_{3} = (3)(5)(7) = 105 = -p_{3} [satisfied.]⇒ roots of g(x) = 0 are 3, 5, 7.

Thus, Solving Polynomials, we get roots of of the given Algebra Equation are 1, 3, 5, 7. Ans.

Example 2 Solving Polynomials is thus solved :

Research-based personalized Math Help tutoring program : Solving Polynomials

Here is a resource for Solid Foundation in Math Fundamentals from Middle thru High School. You can check your self by the

Are you spending lot of money for math tutors to your child and still not satisfied with his/her grades ?

Do you feel that more time from the tutor and more personalized Math Help to identify and fix the problems faced by your child will help ?

Here is a fool proof solution I strongly recommend and that too With a minuscule fraction of the amount you spent on tutors with unconditional 100% money back Guarantee, if you are not satisfied.

It is like having an unlimited time from an excellent Tutor.

It is an Internet-based math tutoring software program that identifies exactly where your child needs help and then creates a personal instruction plan tailored to your child’s specific needs.

If your child can use a computer and access the Internet, he or she can use the program. And your child can access the program anytime from any computer with Internet access.

There is an exclusive, Parent Information Page provides YOU with detailed reports of your child’s progress so you can monitor your child’s success and give them encouragement. These Reports include

Time spent using the program

Assessment results

Personalized remediation curriculum designed for your child

Details the areas of weakness where your child needs additional help

Provides the REASONS WHY your child missed a concept

List of modules accessed and amount of time spent in each module

Quiz results

Creates reports that can be printed and used to discuss issues with your child’s teachers

These reports are created and stored in a secure section of the program, available exclusively to you, the parent. The section is accessed by a password that YOU create and use. No unauthorized users can access this information.

Its research-based results have proven that it really works for all students! in improving math skills and a TWO LETTER GRADE INCREASE in math test scores!,if they invest time in using the program.

Proven for More than 10,000 U.S. public school students who increased their math scores.

Progressive Learning of Math : Solving Polynomials

Recently, I have found a series of math curricula (Both Hard Copy and Digital Copy) developed by a Lady Teacher who taught everyone from Pre-K students to doctoral students and who is a Ph.D. in Mathematics Education.

This series is very different and advantageous over many of the traditional books available. These give students tools that other books do not. Other books just give practice. These teach students “tricks” and new ways to think.

These build a student’s new knowledge of concepts from their existing knowledge. These provide many pages of practice that gradually increases in difficulty and provide constant review.

These also provide teachers and parents with lessons on how to work with the child on the concepts.

The series is low to reasonably priced and include