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Mathematics 42 Online
OpenStudy (anonymous):

z^3+2z^2-z-2 as a product of polynomials of degree 1

OpenStudy (anonymous):

The z^3 term has a coefficient of 1, so the z terms in the factors will have coefficients of 1. The constant term is -2, so the product of the constant terms in the factors will be divisors of -2: -2, -1, 1, or 2. And odd number of them will be negative. Armed with those clues, we can come up with some possible values for one factor, and do long division to see if we were right. Here are the only possible expressions for the factors: x-2 x-1 x+1 x+2 Try each of those until one divides evenly. Then you'll be left with a quadratic expression that you can probably factor by inspection.

OpenStudy (anonymous):

And how can it be solve in form of a linear factor

OpenStudy (anonymous):

Answer is : \[(z-1) (1+z) (2+z)\] (z-1) is a factor and could be verified using remainder theorem

OpenStudy (anonymous):

Well, it's not an equation, so there's no solution. You're just factoring it into 3 factors.

OpenStudy (anonymous):

Remainder theorem. I like that.

OpenStudy (anonymous):

After checking it with remainder theorem we could use Ruffini's rule ( http://en.wikipedia.org/wiki/Ruffini's_rule) or synthetic division for finding the quadratic equation and then the same old quadratic formula to get the desired form.

OpenStudy (anonymous):

If z-1 is used there would be another factor of z^2 +2

OpenStudy (anonymous):

lol. The problem is the example from the page for Ruffini's rule.

OpenStudy (anonymous):

@FoolForMath, what do you know about factoring polynomials with 2 variables, like x^2 - 3xy + 2y^2 - 3x + 5y + 2 Notice the xy and constant terms. The factors will look like (ax+by+c).

OpenStudy (anonymous):

Are you sure you could factor x^2 - 3xy + 2y^2 - 3x + 5y + 2 ? Anyways I guess you meant something like this :\[3 x^3 y + 6 x^2 y^3 + 2 x^3 + 4 x^2 y^2 =x^2 (2+3 y) \left(x+2 y^2\right)\] right ?

OpenStudy (anonymous):

Okay I got it .. sorry . your factor is \[(-1+x-2 y) (-2+x-y)\]

OpenStudy (anonymous):

No. It works out to (x-y-2)(x-2y-1). I spent about an hour working on it, by equating coefficients of like terms. But I was wondering if there was a more sensible/direct way to go about it. You came up with Ruffini's rule and Remainder Theorem, so I thought you might have some more rules floating around.

OpenStudy (anonymous):

How did you get that? By picking a likely factor and testing it by dividing?

OpenStudy (anonymous):

Yes,something like that,I am not sure (at this time) if there is a formal rule,I will get onto you on this later!

OpenStudy (anonymous):

Cool! Thanks.

OpenStudy (anonymous):

Alright there is not simple rule I guess you may try this : http://www.ma.utexas.edu/users/voloch/FFnotes/nov7.pdf

OpenStudy (anonymous):

You da man, Fool! How can I give you more medals?!

OpenStudy (anonymous):

Glad to help :)

OpenStudy (jamesj):

For the record, the easiest way by far to confirm that z - 1 is a factor of the original polynominal is to show that z = 1 is root, which it is.

OpenStudy (anonymous):

@JamesJ: I guess remainder theorem works exactly in that way.

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