Factorising help: x^4 + 3x^3 +x^2

Original post by TenOfThem

ok

the quadratic formula provides solutions to quadratic equations

that is different to factorising

It does the same job, it finds zeroes, factoring finds zeroes. I don't know, I studied EE, but I also studied a lot of math with that, complex roots are important to me as well. Maths is a learning curve, you cant just jump into integration without studying differentiation, and you can't study diff equations before those modules, I may be just jumping the a-level gun, maybe the OP can come back with his teachers opinions. I think we should end this now, I was, after all, only trying to help the guy out.

Reply 42

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Original post by nimrodstower

It does the same job, it finds zeroes, factoring finds zeroes. I don't know, I studied EE, but I also studied a lot of math with that, complex roots are important to me as well. Maths is a learning curve, you cant just jump into integration without studying differentiation, and you can't study diff equations before those modules, I may be just jumping the a-level gun, maybe the OP can come back with his teachers opinions. I think we should end this now, I was, after all, only trying to help the guy out.

I doubt if his teacher's views on this will differ from the views of the maths teachers on this thread

Reply 43

Kallisto

22

To factorise the whole term I would find out the zero points of one:

x^4 + 3x^3+x^2 = 0

. Find the first zero point out by doing, then you should use a polynomial division till the second grade (x²). Then a pq-Formula can be used to get all kinds of zero points of your term. Then you can use the zero points to rewrite the term in linear factors. Then you will get your factoring.

(edited 10 years ago)

Original post by Kallisto

To factorise the whole term I would find out the zero points of one:

x^4 + 3x^3+x^2 = 0

. Find the first zero point out by doing, then you should use a polynomial division till the second grade (x²). Then a pq-Formula can be used to get all kinds of zero points of your term. Then you can use the zero points to rewrite the term in linear factors. Then you will get your factoring.

This thread was done and dusted in post 2 and, if that was too cryptic, post 3 gave the complete answer.

Reply 45

RVNmax

19

tbh I don't understand why nimrodstower is wrong and I'm also studying maths at degree level

Reply 46

Kallisto

Entertainment Forum Helper

22

Original post by Mr M

(...) post 3 gave the complete answer.

I see. Did you mean this post?

Original post by suzylemonade

x^2(x^2 + 3x + 1)
I'm not sure if it's possible to factorize it further :smile:

It was asked for a common factor, not for linear factors, right? Then it was simpler than I thought. My bad.

(edited 10 years ago)

Reply 47

Original post by RVNmax

tbh I don't understand why nimrodstower is wrong and I'm also studying maths at degree level

Well, firstly, factorising a quadratic indicates that you should find the rational factors

Whilst it is true that

x^2-3 = (x+\sqrt3)(x-\sqrt3)

that is not what is being asked when you are asked to factorise

Solve

x^2-3=0

could use those brackets but that is because you are trying to find the roots

And, secondly, he put + when he should have put -

Reply 48

Right I have done a little googling on the matter, and found this:

WIKI HOW:
HOW TO FACTOR SECOND DEGREE POLYNOMIALS.

METHOD 6:

Factor x^2+4x+1; A similar polynomial.

It gives the answer as;

(x+(2+root3)(x+(2-root3):

Go check it out.

Reply 49

Sam Walters

3

Original post by nimrodstower

No body sussed it yet? I hate trying to type math stuff with a keyboard;

(x^2)(x+(-3+root 5)/2)(x+(-3-root 5)/2)

I think that may be readable.

= x^4+1/2 (-3+sqrt(5)) x^3+1/2(-3-sqrt(5)) x^3+1/4((-3-sqrt(5))(-3+sqrt(5))) x^2 (Fair few steps skipped here)
= x^4-3x^3+x^2

So close!

But well. If the lad is being asked a question such as factorise x^4 + 3x^3 +x^2 then x^2(x^2 + 3x + 1) will probably do.

(edited 10 years ago)

Reply 50

Original post by nimrodstower

Right I have done a little googling on the matter, and found this:

WIKI HOW:
HOW TO FACTOR SECOND DEGREE POLYNOMIALS.

METHOD 6:

Factor x^2+4x+1; A similar polynomial.

It gives the answer as;

(x+(2+root3)(x+(2-root3):

Go check it out.

Possibly you were looking at a different page to

http://www.wikihow.com/Factor-Second-Degree-Polynomials-(Quadratic-Equations)

but this page does use the example that you show

However - it is clear from the start that it is looking at the solutions to a quadratic equation

Reply 51

Original post by james22

No idea what level she has done maths to but I'm in the middle of a degree in it and I can say with 100% certainty that you are wrong.

See my post NO. 49, you will see that I am right, and by the way I got my degree in the early eighties.

Reply 52

Original post by nimrodstower

See my post NO. 49, you will see that I am right, and by the way I got my degree in the early eighties.

See my post 50

So did I

Reply 53

Original post by TenOfThem

Possibly you were looking at a different page to

http://www.wikihow.com/Factor-Second-Degree-Polynomials-(Quadratic-Equations)

but this page does use the example that you show

However - it is clear from the start that it is looking at the solutions to a quadratic equation

The page I looked at and which is still on my computer was not about solving but factoring, is it so hard to admit you are wrong? I gave the expression to be factored it contained no =0, so come on admit it you lot are wrong.

Reply 54

Original post by TenOfThem

See my post 50

So did I

t
That's interesting, where? I was a mature student with the OU, I never got the chance to go on to uni from school, young father etc, had to earn some cash. I am retired now, earned it.

Reply 55

Original post by nimrodstower

The page I looked at and which is still on my computer was not about solving but factoring, is it so hard to admit you are wrong? I gave the expression to be factored it contained no =0, so come on admit it you lot are wrong.

*sigh*

As I have said; had you put the - signs in then your expression is indeed equal to the original expression

If trying to find roots it would be perfectly acceptable (though an unnecessary line in the working out)

However - when asked to factorise we do not continue to the point beyond rational factors

I can see that you will not be convinced so it is probably easier to leave this here

Reply 56

Original post by nimrodstower

t
That's interesting, where? I was a mature student with the OU, I never got the chance to go on to uni from school, young father etc, had to earn some cash. I am retired now, earned it.

I read maths at Manchester, straight from school and have been teaching secondary school maths for the intervening 28 years

Reply 57

FireGarden

14

Original post by RVNmax

tbh I don't understand why nimrodstower is wrong and I'm also studying maths at degree level

He's factorising a polynomial with rational coefficients, using real numbers. Aka, cheating.

Supposing he did have a degree, and studied even a little algebra, he would know that any polynomial with real coefficients can be reduced to linear and quadratic factors only, however all polynomials can be factored into only linears over the complex numbers.

This distinction is the same here, though between rationals and reals. He is simply using numbers not allowed in the game. Filthy cheater.

Reply 58