Solve Polynomial: Factoring Tips & Tricks

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    Factoring Polynomial
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any hints on how to start this problem?

$12x^4+19x^3-26x^2-61x-28$
 
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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 $12x^4$?
 
yes that's 12x^4. sorry.
 
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.
 
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...
 
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.
 
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)
 
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]
 
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? :)
 
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. :)