Factoring polynomial

bergausstein

Active member
any hints on how to start this problem?

$12x^4+19x^3-26x^2-61x-28$

Last edited:

Opalg

MHB Oldtimer
Staff member
any hints on how to start this problem?

$12x^2+19x^3-26x^2-61x-28$
Is that $12x^2$ perhaps a typo for $12x^4$?

bergausstein

Active member
yes that's 12x^4. sorry.

Opalg

MHB Oldtimer
Staff member
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.

Deveno

Well-known member
MHB Math Scholar
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.....

LATEBLOOMER

New member
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.

Opalg

MHB Oldtimer
Staff member
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.

soroban

Well-known member
Hello, bergausstein!

Any hints on how to start this problem?

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

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

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

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

$$\text{Can you finish it?}$$

topsquark

Well-known member
MHB Math Helper
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

bergausstein

Active member
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?

LATEBLOOMER

New member
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.