Factoring methods with equations higher than second degree

In summary, there are no general methods for factoring polynomials of higher degrees, but the Rational Roots Theorem can be used to find possible rational roots. If a rational root is found, it can be divided out using synthetic division. If no rational roots are found, other sophisticated methods such as computing modulo a prime and using Hensel lifting can be used. An example of factoring a polynomial using the Rational Roots Theorem is shown, along with a mistake in using synthetic division. A method for solving cubic equations is explained, which involves finding numbers that satisfy a simple quadratic equation.
  • #1
darkmagic
164
0

Homework Statement



Can someone tell what are the other factoring methods when dealing with equations higher than the second degree. The only I is the synthetic division. Aside from that, I do not know anything. I need other factoring methods so I can deal with problems that cannot be factored using synthetic division. I will use them in solving the unknowns in determinants. Please explain briefly to me and give such example.

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
  • #2


There are no general methods of factoring such polynomials. The most helpful thing often is the "rational root theorem". If a polynomial is of the form [itex]a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0[/itex], with all coefficients integers, then any rational number that makes it 0 will be of the form p/q where q is an integer that evenly divides [itex]a_n[/itex], the leading coefficient, and p is an integer that evenly divides [itex]a_0[/itex], the constant term. If [itex]a_n[/itex] and [itex]a_0[/itex] don't have too many factors, you could try all combinations to see if any give a rational root. If there exist a rational root, p/q, then you know that (x- p/q) is a factor. Of course, it might happen that there are NO rational roots and so all factors will have be irrational or complex. In that case, there are not simple ways to find them for degree higher than 2. There exist complicated formulas for degrees 3 and 4 but it was shown in the nineteenth century that polynomial equations of degree 5 and higher may have solutions that cannot be written in terms of roots so there cannot be any formulas, in terms of roots and other algebraic operations, for them.
 
  • #3


so what should I do if the the given cannot be factored by synthetic division? I already encountered one in differential equations and I do not know what to do.
 
  • #4


As HallsofIvy pointed out, there are no general methods, except if you have some additional information (like e.g. you know that the roots are rational). Besides the Rational Roots Theorem, there are then other sophisticated methods like doing computations modulo some prime and then using Hensel lifting to convert that result to modulo some large power of that prime. You can use so-called rational reconstruction methods to convert integers modulo large numbers back to fractions.

https://openaccess.leidenuniv.nl/dspace/handle/1887/3810" for one of the most computationally efficient methods.
 
Last edited by a moderator:
  • #5


an example:
how can you factor
x^3 - 2x^2 + 10x + 7 = 0

It cannot be factored using synthetic division.
 
  • #6


Rational Roots Theorem yields x = -1, 1, 7 and -7 as possible roots. x = -1 turns out to be a root. You then divide the polynomial by x + 1.
 
  • #7


when i used synthetic division, -1 is not a root of that equation. why should I divide the polynomial by x+1?
 
  • #8


darkmagic said:
when i used synthetic division, -1 is not a root of that equation. why should I divide the polynomial by x+1?

Yes, I made a mistake. Well, you can simply solve a qubic equation. First you write it in the form

y^3 + p y + q

by substituting

x = y+2/3

Then you solve that equation by using the identity:

(a + b)^3 = a^3 + 3 a^2b + 3 ab^2 + b^3

You can write this as:

(a+b)^3 = 3ab(a+b) + a^3 + b^3

Now the above equation is an identity, which means that it is always valid no matter what you substitute for a and b. Then, if you want to solve an equation of the form:

y^3 + p y + q = 0

which we can write as:

y^3 = -p y - q

you could try to find numbers a and b such that:

3 a b = -p

a^3 + b^3 = -q

The above identity then implies that y = a + b is a solution.

Finding a and b is easy, if you take the third power of the first equation, you see that you can write bith equation in the form:

A B = - (p/3)^3

A + B = -q

where A = a^3 and B = b^3

Fibnding A and B amounts to solving a simple quadratic equation!
 
  • #9


Can I know what method you used? and teach me to solve this kind of equations?
 

1. What are factoring methods with equations higher than second degree?

Factoring methods with equations higher than second degree refer to the process of breaking down a polynomial equation with a degree higher than two into simpler, smaller expressions that can be multiplied together to get the original equation.

2. Why is factoring important in solving higher degree equations?

Factoring allows us to find the roots or solutions of a polynomial equation, which are the values of the variable that make the equation true. This is useful in many applications, such as finding the x-intercepts of a graph or solving real-world problems.

3. What are the common factoring methods used for equations higher than second degree?

The most common factoring methods include the quadratic formula, grouping, difference of squares, and sum/difference of cubes. These methods vary in their techniques but all aim to simplify the polynomial equation into smaller, more manageable expressions.

4. How do you know when an equation is factorable?

An equation is factorable if it can be written as a product of two or more simpler expressions. This means that the equation has multiple terms and can be reduced through common factors or other factoring methods.

5. Can factoring be used for equations with complex numbers?

Yes, factoring can also be used for equations with complex numbers. Complex numbers are numbers that have both a real and imaginary part, and they can be factored using the same methods as real numbers. The only difference is that the solutions may include imaginary numbers, such as i = √(-1).

Similar threads

  • Calculus and Beyond Homework Help
Replies
9
Views
981
  • Calculus and Beyond Homework Help
Replies
4
Views
650
  • Calculus and Beyond Homework Help
Replies
18
Views
4K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
897
Back
Top