Polynomial / HELM

Sully

Hey,

Can someone guide me to solving a polynomial equation? I know the quadratic one which has a formula, but im unsure on the polynomail type one.

Example: Determine all the solutions of the equation

z^3+3z^2+9z-13=0

Iv searched god knows amount of websites, but I still cant master them. The HELM workbooks have a chapter on it but im unable to locate them online without being asked to authorise myself (college networks and stuff) - so if anyone could point me where to get the HELM workbooks online twould be great!

Thanks lads!

Irish Wolf

Sully wrote:

Hey,

Can someone guide me to solving a polynomial equation? I know the quadratic one which has a formula, but im unsure on the polynomail type one.

Example: Determine all the solutions of the equation

z^3+3z^2+9z-13=0

Iv searched god knows amount of websites, but I still cant master them. The HELM workbooks have a chapter on it but im unable to locate them online without being asked to authorise myself (college networks and stuff) - so if anyone could point me where to get the HELM workbooks online twould be great!

Thanks lads!

:eek: Algebra at this time on a Sunday morning...

Is this what you're looking for? http://www.lboro.ac.uk/research/helm/C_HELM_backup_24nov03/helm_website/documents/wb03_blk3.pdf

Sully

Irish Wolf wrote:

:eek: Algebra at this time on a Sunday morning...

Is this what you're looking for? http://www.lboro.ac.uk/research/helm/C_HELM_backup_24nov03/helm_website/documents/wb03_blk3.pdf

Cheers! Ill take a gawk now. Iv an exam Tuesday morning in Applied Calculus - difficult enough considering im terrible at Maths!

TX123

First byou have to find a common route. for example if u sub in a numer for z and the equation equals zero on both sides then you know this number is a root . e.g if 2 goes into the equation to give zero then z=2. rearrange to get z-2 =0 now using division divide z-2 into the given equation. ur answer is a quadratic equation then just find the roots of that equation.

E.g
X^3+3x^2-4x-12=0
so if we try the factors of 12 we find that subbing -3 into the eqaution gives us zero. so x+3 is a factor now. X^3+3x^2-4x-12 / x+3 we get

x^2-4=0

x^2=4

x=sqrt of 4 =2
so x-2 is the second root

hope that helps

Sully

Erm, slightly confusing - the notes give a different solution!

So we get factors of the last number, and find two that when subbed into the equation give us zero. Thus, giving us two roots. I dont see where it breaks down into a quadratic - is it supposed to?

LeixlipRed

The reason his answer is different is because the last integer he used is 12 not 13 like the question you asked. If you google "factoring a cubic equation" you'll find a formula for factoring all cubic equations

TX123

i found this http://www.sosmath.com/algebra/factor/fac11/fac11.html

it looks complicated not sure how to factor that one because -1,+1,-13,+13 do not make zero ion the equation.

i heard something about the berniscoli method but cant seem to find anything on it

Sully

I found that ages ago, and found it complicated. A friend of mine is going to go over this with me tommrow, she has the solution worked out and knows what to do. If I figure it out, ill post back so its of use to others!

MathsManiac

TX123 wrote:

it looks complicated not sure how to factor that one because -1,+1,-13,+13 do not make zero ion the equation.

Actually, 1 is a root. When you divide out the corresponding factor, which is z-1, the resulting quadratic is z^2 + 4z + 13 = 0, which gives the remaining two roots: -2+3i and -2-3i.

The technique for solving a cubic equation, (provided that it has at least one integer root,) is on the leaving cert course and is well explained in leaving cert textbooks. (They're generally easier to follow than college ones, if the material you need is covered.)

Briefly: suppose in general that the equation is a*x^3 + b*x^2 + c*x + d = 0. Then, IF the equation has an integer root, then that root will be a factor of the number d. (That's why TX123 tried only +1, -1, +13 and -13 as roots in your example; it's just unfortunate that (s)he evidently made a mistake checking 1.) Anyway, that step gets you the first root. After that, you take advantage of the "factor theorem", which tells you that if x=1 is a root of the polynomial, then x-1 is a factor. So, you use algebraic long division to divide x-1 (or z-1 in your case because your equation has z instead of x) into the polynomial. If you started with a cubic expression, you should end up with a quadratic expression when you divide, (and the remainder should be 0; if it isn't, you've made a mistake). Solving that quadratic, using the formula or otherwise, gives you the other two roots.

There are a number of techniques for moving beyond that level of difficulty, (that is, for dealing with polynomials with powers higher than 3, or for dealing with cubics that don't have an integer root) but your message doesn't make it clear how much you need to know. Post one or two examples from past papers of the hardest kind you need to deal with, and I'll try to help (as will others I'm sure).

LeixlipRed

If you're really struggling you could just learn off the general formula as i suggested above. As seen here. Of course a question on an exam paper will nearly always involve the method described above by MathsManiac and that formula isn't the easiest thing to remember either

seandoiler

MathsManiac wrote:

Briefly: suppose in general that the equation is a*x^3 + b*x^2 + c*x + d = 0. Then, IF the equation has an integer root, then that root will be a factor of the number d.

strictly speaking this is not true for the equation you have written unless a=1, if you have an integer root for a not equal to 1, the root would have to be a factor of (d/a) not just d...cf Gauss' Lemma

MathsManiac

seandoiler wrote:

strictly speaking this is not true for the equation you have written unless a=1, if you have an integer root for a not equal to 1, the root would have to be a factor of (d/a) not just d...cf Gauss' Lemma

Indeed, I was aware of that, but trying to keept things simple, so as to possibly remind the OP of a technique they might have used at school.

Since the root is a factor of (d/a), then it will also be a factor of d, so the assertion was still true. It is more helpful of course to know that it's a factor of d/a, but you're incorrect to make the assertion that "this is not true for...".

If the OP indicates more techniques are required, one can expand.

(I'm off on a week's hols now, so going silent!)

seandoiler

well it is true for integer coefficients and integer roots but if you take the coefficient a as a fraction (contrived, I know)...but in this case the situation may arise that the factor divides (d/a) but not d...more specifically i was thinking about rational as opposed to integer roots for the general case...anyway i think in most cases that people will encounter, you are right and i probably shouldn't have complicated things

Sully

hey,

Its in the Leaving Cert, and iv been showen what to..

Get a root, confirm by making sure it brings it to 0, divide x-r (where r is the root) into the polynomial to get a quadratic. From there, the roots are exposed by using the -b formula for quadratics.

Make sense?

seandoiler

Sully wrote:

Make sense?

this is the method that was explained in the topic a few posts earlier by Mathsmaniac

Sully

I got one of these in an exam and was unable to find a root, some people approached it using the quadratic -b formula tho - even tho it wasnt layed out like one.

Ah sure, thats life! Thanks

LeixlipRed

Have you got the question there?

MathsManiac

seandoiler wrote:

well it is true for integer coefficients and integer roots but if you take the coefficient a as a fraction (contrived, I know)...but in this case the situation may arise that the factor divides (d/a) but not d...more specifically i was thinking about rational as opposed to integer roots for the general case...anyway i think in most cases that people will encounter, you are right and i probably shouldn't have complicated things

Just got back from hols, so I know the thread has petered out a bit, but I just wanted to point out that you obviously have to assume the coefficients to be integers so that the discussion of divisibility makes any sense. If you are considering rational coefficients, then the divisibility lemmas are meaningless, since, in the rationals, every number divides every other number (provided you're careful about 0). Hence, as I pointed out, an integer root can't divide d/a without also dividing d.

Even when extending this technique to the standard result regarding the condition on a rational root, (numerator divides d and denominator divides a) one still considers the polynomial to have integer coefficients, as otherwise divisibility is meaningless.

And, since every polynomial over the rationals is equivalent (as far as its roots are concerned) to a polynomial over the integers, you lose nothing by dealing with the case of integer coefficients only, which is why people generally do this.

seandoiler

MathsManiac wrote:

Just got back from hols, so I know the thread has petered out a bit, but I just wanted to point out that you obviously have to assume the coefficients to be integers so that the discussion of divisibility makes any sense.

of course, sorry never thought about it that way.....but as i tried to say i was more concerned with rational roots, for example in the eq 3x^3 - 2x^2 + 12x - 8 which has root 2/3...just wanted to point out the more general case of gauss' lemma...but i made a mess of it