Maybe factoring a polynomial completely

[Perhaps check it again]

Can the following be completely factored over integer coefficients, where the highest degrees

on x and y are both one?

 

 

2x²  + 3xy - 2y² - x + 3y - 1

 

 

If not, state that.

 

If so, then show the steps leading up to and the final factored form.

[BMAD]

ac = 2

ad + bc = 3

bd = -2

c - a = -1 

d - b = 3

 

d = 3 + b

b(3+b) = -2

b²+3b+2 = 0

(b+2)(b+1) = 0, b = -2 and d = 1 or b =-1 and d=2

 

c - a = -1 

c = -1 + a

ac = 2

a(-1+a)=2

a²-a-2=0

(a-2)(a+1)=0; a=2 and c =1 or a=-1 and c=-2

 

b = -2 and d = 1 or b =-1 and d=2

 a=2 and c =1 or a=-1 and c=-2

 

ad = 4 or 2 or -1 or -2

bc = -2 or 4 or -1 or 2

 

given:

ad + bc = 3

 

ad = 4 when bc = -1 so a = 2, d = 2, b = -1, c = 1

ad= -1 when bc = 4 so a = -1, d = 1, b = -2, c = -2

 

there are two answers

[harey]

a) -1 on end means (...+1)(....-1)
b) as there is no term with power 3, I made the bet there only are 2 terms

c) multiplied (ax+by+1)(cx+dy-1)

d) ac=2; (c-a)=1; I would find very tasteful a=2; c=1;

e) bd=-2; (d-b)=3; b=-1 and d=2 ???

f) (ad+bc)=(2*2-1)=3: It works!!!

FInal form:(2x-y+1)(x+2y-1)

 

Edited January 12, 2015 by harey

[Perhaps check it again]

Regarding post #2, you have a certain amount of doubt regarding correctness and/or

justification of certain steps, as evidenced by question marks (as a for instance).

 

 

I am looking for a solution by anyone made with complete confidence, and where

the steps are justified from the prior ones.  I'm not wanting any steps that come about

through luck/inspection/trial and error.

Edited January 12, 2015 by Perhaps check it again

[harey]

Well, I am just not very sure concerning the assumption in b). I do not consider whether there is a solution of another kind - when solving a problem, there is a little bit of intuition on the beginning. This does not make me doubt about the solution if I find one, except in the case I would claim it is unique.

 

As for the remainder, there are so few possible solutions (we look for integers) that you can try them all. Otherwise, you could solve it as a system of multiple equations with multiple variables, but I think it is too much hassle here.

 

I am awaiting the official solution!!!

[Perhaps check it again]

harey,

 

if you'll just finish the problem, you've got it.

 

The problem does not have to be solved by a system of equations, but because

your post involved that, I am posting continuing work under this spoiler for you.

 

Here is a system of simultaneous equations: 

 

Multiply out your factors to:

 

(ac)x² + (ad + bc)xy + (bd)y² + (c - a)x + (d - b)y - 1

 

ac = 2

 

ad + bc = 3

 

bd = -2

 

c - a = -1    You mistakenly have this as positive one in your post.

 

d - b = 3

 

 

Let's just agree to choose positive coefficients for a and c in the answer.

(I know you already posted the final solution.)

 

I'm looking for you to show the steps in solving the system of simultaneous

equations above.

[harey]

c - a = -1    You mistakenly have this as positive one in your post.

 

This is just a typo, sorry for that:

d) ac=2; (c-a)=1; I would find very tasteful a=2; c=1;

(1-2)=-1

 

 

I'm looking for you to show the steps in solving the system of simultaneous equations above.

 

Sorry, I fear I will disappoint you (as I already said, there are few possibilities, so a brute force approach will solve the system in seconds while I would spend hours due to algebra errors):

 

for a in [-2,-1,1,2]:
  c=int(2/a)
  for b in [-2,-1,1,2]:
    d=int(-2/b)
    if(d-b==3):
      if(a*d+b*c==3):
        print("a=",a,"c=",c,"b=",b,"d=",d,"ad+bc=",a*d+b*c,"d-b=",d-b)
        print(a*c,"x2+",a*d+b*c,"xy+",b*d,"y2+",c-a,"x+",d-b,"y-1")

 

 

 

The problem does not have to be solved by a system of equations,

As no one else posts and I have no other idea how to solve it, can you post the solution?

[DejMar]

 

A polynomial of the form (ac)x

² + (ad + bc)xy + (bd)y² + (an + mc)x + (bn + md)y + mn can be expressed as having the factors
(ax + by + m) and (cx + dy + n).

Equating the coefficients of the polynomial 2x² + 3xy - 2y² - x + 3y - 1 to the expression above:
A:  2 = ac
B:  3 = ad + bc
C: -2 = bd
D: 1 = an + mc
E: 3 = md + bn
F: -1 = mn

WLOG, and seeking only integer coefficients, let m = 1, n = -1
from (E) 3 = d - b, thus, d = b + 3
substituting into ( C),
-2 = b(b + 3)
0 = b² +3b + 2, which factors to
0 = (b + 2)(b + 1)
b = {-1, -2}
therefore, by again substituting into © and solving
d = {2, 1}

b = -1, d = 2         b = -2, d = 1
3 = a(2) + (-1)c      3 = a(1) + (-2)c
3 = 2a - c            3 = a - 2c
substituting (A), a = 2/c,
3 = 2(2/c) - c        3 = (2/c) - 2c
multiplying both sides by c and factoring
c² + 3c - 4 = 0      2c² + 3c - 2 = 0
(c + 4)(c - 1) = 0    (2c - 1)(c + 2) = 0
c = {-4, 1}           c = {1/2, -2}
therefore,
2 = a(-4)  2 = a(1)   2 = a(1/2)  2 = a(-2)    
a = -1/2   a = 2      a = 4       a = -1
Ignoring the non-integer coefficients, the factors are then
G: (2x - y + 1)(x + 2y - 1), or
H: (-x - 2y + 1)(-2x + y - 1)

(H) is the same as (G) but with each factor multiplied by -1, thus
the factors for the equation can be simply given as
(2x - y + 1), (x + 2y - 1)

Edited January 26, 2015 by DejMar