dagesi
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How do I factor polynomials
Ok, I'm trying to teach my kid how to factor polynomials. I already know how to factor bi- and tri- nomials and I already know how to take the GCF (Greatest Common Factor), for instance:
With the equation:
x^3 - 7x^2 - 16x + 112
This can be grouped into (x^3 - 7x^2) - (16x -112)
Then can factor out:
x^2(x - 7) - 16(x - 7)
Which then becomes:
(x^2 - 16)(x - 7)
And finally:
(x - 4)(x + 4)(x - 7)
That's all well and good... but what about something like:
x^3 + 3x^2 - 46x - 168
In this case, if I KNOW any of the factors already, I can determine the rest... but how do I get a factor to begin with? The GFC doesn't seem to work for this...
And, again, I want to point out this is me trying to learn how to teach someone else, not actual homework...
(And to demonstrate that I know the answer):
(x+ 6)(x - 7)(x + 4)
With the equation:
x^3 - 7x^2 - 16x + 112
This can be grouped into (x^3 - 7x^2) - (16x -112)
Then can factor out:
x^2(x - 7) - 16(x - 7)
Which then becomes:
(x^2 - 16)(x - 7)
And finally:
(x - 4)(x + 4)(x - 7)
That's all well and good... but what about something like:
x^3 + 3x^2 - 46x - 168
In this case, if I KNOW any of the factors already, I can determine the rest... but how do I get a factor to begin with? The GFC doesn't seem to work for this...
And, again, I want to point out this is me trying to learn how to teach someone else, not actual homework...
(And to demonstrate that I know the answer):
(x+ 6)(x - 7)(x + 4)
ASKER
>ozo...
I guess when I said "trying to teach my kid" I should have mentioned he's 14...
I don't even understand the numerical method (what level of math is that?)...
As for the analytical method, do you mean that that is the EASY way of doing it? Meaning there isn't a method as simple as there is for the bi- or tri-...?
I'm still working through that method to see if I understand it...
(Boy, I feel stupid...) =]
I guess when I said "trying to teach my kid" I should have mentioned he's 14...
I don't even understand the numerical method (what level of math is that?)...
As for the analytical method, do you mean that that is the EASY way of doing it? Meaning there isn't a method as simple as there is for the bi- or tri-...?
I'm still working through that method to see if I understand it...
(Boy, I feel stupid...) =]
SOLUTION
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ASKER
>Infinity...
I actually already saw that page while searching for a method...
I knew about (and explained to) the method where you trial and error using factors, ie:
6x^2 + 23x + 20
6 can be 1 & 6, 2 & 3 or vice versa or their negatives...
20 can be 1 & 20, 2 & 10, 4 & 5 or vice versa or their negatives...
Trial and error each different combination if needed (or as skill improves, automatically ignore some)...
So, no guaranteed method like that for finding the answer in this case, right...?
The reason the Q came up for me is I saw some problems coming up for him that provide one or two factors and ask you to determine the rest - I was expecting there might be a next step where none was provided and some relatively easy (and new) method might have to be learned to figure all of them...
I actually already saw that page while searching for a method...
I knew about (and explained to) the method where you trial and error using factors, ie:
6x^2 + 23x + 20
6 can be 1 & 6, 2 & 3 or vice versa or their negatives...
20 can be 1 & 20, 2 & 10, 4 & 5 or vice versa or their negatives...
Trial and error each different combination if needed (or as skill improves, automatically ignore some)...
So, no guaranteed method like that for finding the answer in this case, right...?
The reason the Q came up for me is I saw some problems coming up for him that provide one or two factors and ask you to determine the rest - I was expecting there might be a next step where none was provided and some relatively easy (and new) method might have to be learned to figure all of them...
>> So, no guaranteed method like that for finding the answer in this case, right...?
Well, there are the links ozo posted. But they're a bit more advanced and more complicated (too complicated for this kind of simple factoring exercise imo).
>> I was expecting there might be a next step where none was provided
That seems likely, yes. Unfortunately, the "guess" method works best in exercises like these, especially if you practice a lot (and thus gain experience, and can pretty much "see" the correct solution).
Well, there are the links ozo posted. But they're a bit more advanced and more complicated (too complicated for this kind of simple factoring exercise imo).
>> I was expecting there might be a next step where none was provided
That seems likely, yes. Unfortunately, the "guess" method works best in exercises like these, especially if you practice a lot (and thus gain experience, and can pretty much "see" the correct solution).
ASKER CERTIFIED SOLUTION
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in this case 6*7*4 = 168
> provide one or two factors and ask you to determine the rest
Dividing the polynomial by the provided factors would give you a simpler polynomial to factor the rest
Dividing the polynomial by the provided factors would give you a simpler polynomial to factor the rest
SOLUTION
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> now try to look for 3 numbers and with + and - signs to reach 3 (coef' of X^2)
An alternative to looking for all 3 numbers at once is to look for the numbers one at a time.
If any one of the numbers is a root, you can remove it and be left with a simpler polynomial to find the remaining roots.
(And once you reduce it to a quadratic, the quadratic formula is much simpler than the cubic formula)
An alternative to looking for all 3 numbers at once is to look for the numbers one at a time.
If any one of the numbers is a root, you can remove it and be left with a simpler polynomial to find the remaining roots.
(And once you reduce it to a quadratic, the quadratic formula is much simpler than the cubic formula)
since a cubic is guaranteed to have at least one real root between a sufficiently large number and a sufficiently small number
you can always use this method
http://en.wikipedia.org/wiki/Bisection_method
If you only want integer roots, you can stop when your interval size becomes less than 1
you can always use this method
http://en.wikipedia.org/wiki/Bisection_method
If you only want integer roots, you can stop when your interval size becomes less than 1
ASKER
Ok, think I've got it... I'll have to "do some tests of it" but some of these almost make sense... =]
thx...
thx...
ASKER
Going to work on this for a while to be sure I've got it... hopefully I'll understand it before I need to... =]
http://mathworld.wolfram.com/NewtonsMethod.html
analytically
http://www.dpmms.cam.ac.uk/~wtg10/cubic.html