Solve Polynomial: Factoring Tips & Tricks

In summary, the problem may have been misspelled and the simplest solution is to find integer roots of the polynomial. First, look for factors of the constant term and if you find any, then the factor theorem will give you linear divisors of the polynomial.
  • #1
bergausstein
191
0
any hints on how to start this problem?

$12x^4+19x^3-26x^2-61x-28$
 
Last edited:
Mathematics news on Phys.org
  • #2
bergausstein said:
any hints on how to start this problem?

$12x^2+19x^3-26x^2-61x-28$
Is that $12x^2$ perhaps a typo for $12x^4$?
 
  • #3
yes that's 12x^4. sorry.
 
  • #4
Start by looking for integer roots of the polynomial (factors of the constant term). If you find any, then the factor theorem gives you linear divisors of the polynomial.
 
  • #5
Opalg said:
Start by looking for integer roots of the polynomial (factors of the constant term). If you find any, then the factor theorem gives you linear divisors of the polynomial.
The roots may not be INTEGERS, as the leading term's coefficient is not 1...
 
  • #6
i will use dorobostikerlines method.,

$12(x^2-1)^2+19(x^2-1)(x+1)-21(x+1)^2$

i'll let you continue.
 
  • #7
Deveno said:
The roots may not be INTEGERS, as the leading term's coefficient is not 1...
True, but I like an easy life, so I look for the simplest possible solutions first. (Wink)
 
  • #8
Hello, bergausstein!

Any hints on how to start this problem?

$\text{Factor: }\:f(x) \:=\:12x^4+19x^3-26x^2-61x-28$

[tex]\text{We find that }f(\text{-}1) \,=\,0.[/tex]
[tex]\text{Hence, }x+1\text{ is a factor.}[/tex]

[tex]\text{Long division: }\:f(x) \:=\: (x+1)\underbrace{(12x^3 + 7x^2 - 33x - 28)}_{g(x)}[/tex]
[tex]\text{We find that }g(\text{-}1) \,=\,0.[/tex]
[tex]\text{Hence, }x+1\text{ is a factor.}[/tex]

[tex]\text{Long division: }\:g(x) \:=\: (x+1)(12x^2-5x - 28)[/tex]

[tex]\text{Can you finish it?}[/tex]
 
  • #9
LATEBLOOMER said:
i will use dorobostikerlines method.,

$12(x^2-1)^2+19(x^2-1)(x+1)-21(x+1)^2$

i'll let you continue.
I've never heard of this method and google comes up with nothing. Can you give us a quick run-down?

-Dan
 
  • #10
latebloomer, that method seems to work correctly. and i got the right answer. can you show me how that method work in its full glory? :)
 
  • #11
actually you won't like it if I show you the full workings of this method. that method is from an indian mathematician named doroboski. well, his name was not celebrated as other great mathematicians out there so you'll rarely find information about him. :) i would prefer using the other method metioned above. :)
 

Related to Solve Polynomial: Factoring Tips & Tricks

1. What is factoring a polynomial?

Factoring a polynomial means breaking it down into simpler forms, such as multiplying two or more polynomials to get the original polynomial. It is a useful tool in solving equations and understanding the behavior of polynomial functions.

2. Why is factoring important in mathematics?

Factoring is important in mathematics because it helps simplify complicated expressions and equations, making them easier to solve. It also allows us to find the roots or zeros of a polynomial, which are important in many applications, such as graphing and optimization problems.

3. What are some tips for factoring polynomials?

Some tips for factoring polynomials include looking for common factors, using the distributive property, and grouping terms with common variables. It is also helpful to check for special patterns, such as the difference of squares or perfect square trinomials.

4. How do I know if a polynomial can be factored?

A polynomial can be factored if it has more than one term and if it is not a prime polynomial (meaning it cannot be factored further). Additionally, some polynomials may require advanced factoring techniques, such as the quadratic formula or completing the square.

5. Can factoring be used to solve all polynomial equations?

No, factoring can only be used to solve certain types of polynomial equations, such as quadratic equations or equations with special patterns. For more complicated equations, other methods such as the quadratic formula or graphing may be necessary.

Similar threads

  • General Math
Replies
5
Views
1K
Replies
1
Views
704
Replies
3
Views
1K
Replies
2
Views
860
  • General Math
Replies
7
Views
927
Replies
1
Views
866
  • General Math
Replies
1
Views
1K
Replies
3
Views
862
Replies
3
Views
1K
Replies
3
Views
1K
Back
Top