Factorising polynomials (1 Viewer)

[klaw]

When factorising something with a multiple root, e.g. P(x)=(x-@)^3 Q(x), is there a fast way to do it (e.g. by inspection)? Dividing the poly by (x-@)^3 or expanding it and equating coeffs can take a long time. I can find the first and last term of the factor by inspection, but I have to expand it and equate to find the terms in between, e.g. for something like P(x)=(x-@)^3(ax^3+bx^2+cx+d), I can find a and d almost straight away but I have to find b and c using a long way.

 

[Riviet]

Whoah, already on polynomials eh? From my knowledge of this topic, there are heaps of cool tricks on factorising polynomials, so there probably is some efficient way (there better be)...

 

[VivianHsu]

"When factorising something with a multiple root, e.g. P(x)=(x-@)^3 Q(x), is there a fast way to do it (e.g. by inspection)?"

Not really...
you can put x=1 and x=2 into P(x)=(x-@)^3(ax^3+bx^2+cx+d), but you still need to solve simultaneous equations.

 

[SeDaTeD]

It is a theorem that you will learn later that if P(x) is a polynomail with a root @ with multiplicity n, then P'(x) will be a polynomial with the root @ of multiplicity n-1. This may or may not be faster than expanding and collecting.

Proof, let P(x) = a(x-@)^n*Q(x), where Q(x) is another polynomial of degree deg(P(x)) - n.
then P'(x) = a(x-@)^n*Q'(x) + an(x-@)^(n-1)*Q(x)
= (x-@)^(n-1)[a(x-@)Q'(x) + anQ(x)]

therefore @ is a root of multiplicity n-1 of P'(x)

this can also be generalised to higher degrees o multiplicity, ie keep differentiating.

I also know a method where when i know a few roots i can rewrite the original P(x) by adding and subtracting terms so that things will factorise out. Pretty much polynomial division in 2 lines.

 

[airie]

yike i totally forgot what is multiplicity again...>.< gotta do work :S

 

[Trebla]

That inspection method is Terry Lee's cheating method for factorising. lol

 

[SeDaTeD]

haha, I don;t think u can say its exclusively terry lee's method, if you are refering to what i said. I figured that out myself.

Oh and airie, i don't think you need to worry that you don;t know multiplicity, considering you are doing your hsc in 2007. But anyway, multiplicity just means how many times a number is a root of a polynomial (roughly speaking). eg double roots have multiplicity 2, and triple roots, multiplicty 3 and so on.

 

[Riviet]

That inspection method is Terry Lee's cheating method for factorising. lol

Yes! I got his book and his trick is probably in there... hehe. I'll check it out and see.

 

[SeDaTeD]

Why do you refer to it as 'his'? haha

 

[klaw]

I also know a method where when i know a few roots i can rewrite the original P(x) by adding and subtracting terms so that things will factorise out. Pretty much polynomial division in 2 lines.

I don't get you Can you give me an e.g.?

 

[Riviet]

Yeah, give us an example! :uhhuh:

 

[SeDaTeD]

Ok, consider P(x) = x^4 - 3x^3 - 35x^2 + 39x + 70
once you found some factors suchs as -1 and 2, by
P(-1) = 1 + 3 - 35 - 39 + 70 = 0 and P(2) = 16 -24 -140 + 78 + 70 = 0, this can be done on calculator to check.
therefore x+1 and x-2 are factors, therefore x^2 -x -2 is a factor
so P(x) = x^4 - 3x^3 - 35x^2 + 39x + 70
= x^4 - x^3 - 2x^2 - 2x^3 + 2x^2 + 4x -35x^2 + 35x + 70 (**)
= (x^4 - x^3 - 2x^2) + (- 2x^3 + 2x^2 + 4x) + (-35x^2 + 35x + 70) ... I'm just putting brackets here for clarity
= x^2(x^2 -x -2) -2x(x^2 -x -2) - 35(x^2 -x -2)
= (x^2 -x -2)(x^2 -2x - 35)
= (x-2)(x+1)(x+5)(x-7)

(**) what i have done here is to break up the terms into mulitples of x^2 -x -2, ie, since it starts with x^4 i make the terms x^4 - x^3 - 2x^2 whihc are x^2 times x^2 -x -2. Then i move on to x^3 and put a multiple of it that would satisfy the equality, ie ive got -x^3 from the bit before, and all the x^3 terms must sum up to - 3x^3, given by P(x), so i must make the next term -2x^3. Then i continue and add the x^2 and x terms which would make it a multiple of x^2 -x -2, then so on. You now what terms to add because they must sum up to the original P(x).

This is in fact the same process as polynomial division, where you would divide P(x) by x^2 -x -2, but a much bigger space saver.

Normally when i'd write that, i'd just use lines 1, 2, 5 and 6. Maybe not 2 lines but still.

Also, if you know a couple of roots, it is sometimes easier to use the relationships between the roots and the coefficients of terms (eg sum of roots, products of roots etc).

 

[pLuvia]

Aaa, that's pretty good, do you learn that in 4u polynomials?

 

[SeDaTeD]

Nope, they never taught us that.
I came up with the method myself. When you are writing it down maybe leave a little gap between every 3 terms (or what number u need) or put backets in later so at least hte marker can see what you are doing. It's also fine to spread it over a few lines, might be easier to follow and find mistakes.