Can someone refresh my memory on how to factor quartics

ggeek101 · #1 $Old$ November 5th, 2009, 07:54 PM

I need to find all of the zeros.

g(x) = x^4 + x^3 -2x^2 +6x -4

Bacterius · #2 $Old$ November 5th, 2009, 08:19 PM

This is not a quadratic but a polynomial of the fourth degree (with at most 4 solutions).
I believe it is quite hard to factorize and/or solve fourth degree polynomials ?

ggeek101 · #3 $Old$ November 5th, 2009, 08:29 PM

This is a problem on my homework for tonight. I have tried a few ways to solve it; none of them working I am at a loss. Perhaps it would be best to just ask in class. Thanks anyways, though.

Bacterius · #4 $Old$ November 5th, 2009, 08:33 PM

Oh sh** sorry I thought you had written "quadratics", but you indeed wrote "quartics". I'm out, too hard for me ...
But maybe Wikipedia could help ?

http://en.wikipedia.org/wiki/Quartic_function

Wilmer · #5 $Old$ November 5th, 2009, 09:01 PM

Quote:

Originally Posted by ggeek101 $View Post$

I need to find all of the zeros.
g(x) = x^4 + x^3 -2x^2 +6x -4

Are you sure about "+ x^3"? Could that be "- x^3"?
If so, much easier, since divisible by (x - 1): (x - 1)(x^3 - 2x +4)

TheEmptySet · #6 $Old$ November 5th, 2009, 09:11 PM

Quote:

Originally Posted by ggeek101 $View Post$

I need to find all of the zeros.

g(x) = x^4 + x^3 -2x^2 +6x -4

So the bad new is that is has no linear factors. We can check for linear factors by using the rational roots theorem.

So It must factor into two quadratics

since the lead term has coeffient 1 we can write the two quadratics as

$(x^2+ax+b)\cdot (x^2+mx+n) =x^4+x^3-2x^2+6x-4$

Now we multiply out the left hand side to get

$x^4+(m+a)x^3+(n+am+b)x^2+(an+bm)x+bn=x^4+x^3-2x^2+6x-4$

This give the system of equations that is hard to solve and requires alot of trial and error.

$m+a=1,n+am+b=-2,an+bm=6,bn=-4$

Now comes the hard work solving this system of equations.

If $b=2,n=-2$ then if we plug into the 2nd equation

$am=-2$ this gives two choices for a and m

The correct choice is $a=-1,m=2$ they both check in equation 3.

Now plugging back into the equation from the beginning we get
$(x^2-x+2)\cdot (x^2+2x-2)$

artvandalay11 · #7 $Old$ November 5th, 2009, 09:18 PM

Quote:

Originally Posted by ggeek101 $View Post$

I need to find all of the zeros.

g(x) = x^4 + x^3 -2x^2 +6x -4

It's impossible to factor quartics... just kidding, but I can't offer you a better method than guess and check, I mean I could, but I don't think it's any better, so....

Quartics can be factored into two quadractics (provided they have real coefficients)

$g(x) = x^4 + x^3 -2x^2 +6x -4$

$g(x)=(x^2+\quad )(x^2+\quad )$

Now I start playing around with stuff, we need a $+x^3$ and a $-2x^2$

Well it looks like $-x$ and $+2x$ will do the job
Since $x^2(-x)+x^2(2x)=x^3$ and $(-x)(2x^2)=-2x^2$

So $g(x)=(x^2-x+\quad )(x^2+2x+\quad )$

Now the final number is -4, so our choices consist of 1,2,4 with necessary signs out in front, just running through those, again I swear this is all guess and check and it's not necessarily quick, you'll get 2 and -2

$g(x)=(x^2-x+2)(x^2+2x-2)$

There are formulas and things you can look up and use, but this question comes up so rarely for me that I have to stumble through it. It's not an efficient way of doing it.

ggeek101 · #8 $Old$ November 6th, 2009, 03:52 AM

Quote:

Originally Posted by Wilmer $View Post$

Are you sure about "+ x^3"? Could that be "- x^3"?
If so, much easier, since divisible by (x - 1): (x - 1)(x^3 - 2x +4)

Yes, unfortunately. I am sure it is +x^3, but it is possible my instructor made a mistake.