Question

How to factorize a large multinomial ?

Answer 1

like :
\[\Large P(x,y)= x^2 - xy - 2y^2 + 2x - 4y\]
need a good and fast method :(

Answer 2

Notice here that \(x = 2y\) makes \(P(x, y) = 0\), which means that \(x - 2y \) is a factor.

Answer 3

In general, pata nahi... I just try to note things like that ^^^

Answer 4

hmmn :| here, we can break 2y^2 into y^2 + y^2 and then factor it but its about luck, you might be lucky if you spot the right thing. But I am looking for something concrete, if it exists :|

Answer 5

@ganeshie8

Answer 6

Let's take a better question..
x^2+y^2-2xy-3x+3y+2

Answer 7

ah, pair of straight lines

Answer 8

To me, treat x as variable and y as constant

Answer 9

\(x^2-(2y+3)x +(y^2+3y+2)\)
then we have a quadratic with a =1, b= -(2y+3) and c = y^2+3y+2
Hence, we can apply quadratic formula to find the roots --> factor is doable.

Answer 10

this can get ugly at times..

Answer 11

hahaha... you said you want a concrete method, right?

Answer 12

yep..something that works always D: and is easy/less time consuming

Answer 13

ugly or not doesn't matter, as long as you can factor, you are OK>

Answer 14

if the answer is simple, then it won't get ugly.
and ugh how I could I forget this one

Answer 15

Actually, discriminant =1, beautiful enough. :)

Answer 16

That actually works very nicely
We can factor it by picking two binomials such that they addup to the mid term and multiply to the last term

Answer 17

Try y+2 and y+1

Answer 18

yup, but not try!! that is the result from quadratic method.

Answer 19

alright..how do we proceed from here btw ?
\[\Large \frac{(2y+3) \pm \sqrt{(1-24y)}}{2}\]

Answer 20

I think we don't need to know quadratic formula to do this

Answer 21

Remember factoring quadratics by splitting the middle term ?

Answer 22

@ganeshie8 he said that he want a concrete method to apply to any problem, not a particular one and no luck.

Answer 23

yes..I do..but again its bruteforce for finding 2 such binomials, but I guess it should have a better probability of finding a solution

Answer 24

Alright I just thought that quadratic formula is a bit boring

Answer 25

but how do we proceed after that ? :| do I need to make a perfect square ?
anyways, quadratic formula + bit of binomial guessing should do it tho.

Answer 26

The thing under radical should be 1
Double check...

Answer 27

yep..right..sorry (:

Answer 28

thanks everyone!! :D

Answer 29

Since when has factoring had a concrete method?

Even with finding two numbers that add up to the middle term and multiply to the last term... you're still using trial and error.

Answer 30

You could also compare the coefficients both sides and solve the resulting eqns.

Answer 31

I didn't mean exactly concrete, but something which is faster than normal hit and trial for large polynomials

Answer 32

Suppose the given quadratic factors to

(x+ay+b)(x+cy)

Answer 33

We can't be sure about the form of the final equation but ?

Answer 34

Set that equal to the given quadratic
Compare coefficients

Answer 35

We can be sure. Its not hard to see that there is no constant term in the given quadratic
So...

Answer 36

yeah..that'd work..+1