- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,920

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,920

- Jan 26, 2012

- 644

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

(padding above to hide spoiler from "New Forum Posts")

$$(x-2 y) (2 x+3 y-5) (2 x+3 y-1) = 0$$

Which gives us the following conditions:

$$x = 2y ~ ~ ~ ~ \text{and} ~ ~ ~ ~ 2x + 3y = 5 ~ ~ ~ ~ \text{and} ~ ~ ~ ~ 2x + 3y = 1$$

[JUSTIFY]Note that since we only want positive integer solutions, the second condition yields only $(1, 1)$ and the last one clearly cannot be satisfied in $\mathbb{N}$, therefore the set of solutions $(x, y)$ over the positive integers, is in fact equal to the following:[/JUSTIFY]

$$\{ (2n, n) ~ | ~ n \in \mathbb{N} \} ~ \cup ~ \{ (1, 1) \} ~ ~ = ~ ~ \{ (1, 1), \, (2, 1), \, (4, 2), \, (6, 3), \, \dots \}$$

You really have to see the factorisation, though.

- Thread starter
- Admin
- #3

- Feb 14, 2012

- 3,920

Thanks for participating in this problem.

I've come up with the same product of three factors as another way to express the given equation but I'm really more interested in the method that you used to find those factors...

- Jan 26, 2012

- 644

[JUSTIFY]I noticed the pattern with the $x = 2y$ solutions, and tried to divide the polynomial by $(x - 2y)$, and after some rearranging I got a quadratic in $(2x + 3y)$, which wasn't too hard to factor.

Thanks for participating in this problem.

I've come up with the same product of three factors as another way to express the given equation but I'm really more interested in the method that you used to find those factors...

Without having an "insight" into the solutions, I think it would still be doable by analyzing the coefficients...[/JUSTIFY]

- Mar 31, 2013

- 1,349

I found this problem recently and as this is not solved to the extent of factorization I thought I can give it a try[JUSTIFY]I noticed the pattern with the $x = 2y$ solutions, and tried to divide the polynomial by $(x - 2y)$, and after some rearranging I got a quadratic in $(2x + 3y)$, which wasn't too hard to factor.

Without having an "insight" into the solutions, I think it would still be doable by analyzing the coefficients...[/JUSTIFY]

4x^3−12x^2 +4x^2y +5x+6xy−15xy^2−10y+36y^2−18y^3

Now factor the part independent of x

= 4x^3−12x^2 +4x^2y +5x+6xy−15xy^2−2y(5-18y+9y^2)

= 4x^3−12x^2 +4x^2y +5x+6xy−15xy^2−2y(3y-5)(3y-1)

Now I multiply by 2 and put 2x = z to get coefficient of x^3 as 1

= ½(8x^3−24x^2 +8x^2y +10x+12xy−30xy^2−4y(3y-5)(3y-1)

= 1/2(z^3−6z^2 +2 z^2y +5z+6zy−15zy^2−4y(3y-5)(3y-1)

= 1/2(z^3−z^2(6-2y) +z(5+6y+15y^2) −4y(3y-5)(3y-1)

Now we need to split - 4y(3y-5)(3y-1)into3 parts that the sum is 6- 2y

They are-( 3y-5),- (3y-1) , 4y ( it is easy to do so as 5 + 1 = 6

Now to check The coefficient of z is (5+6y+15y^2)

We see that – 4y(3y-5) – 4y(3y-1) + (3y-1) (3y-5) = -15y^2 + 6y + 5 which is true

So we get ½(( z- 4y) ( z+ 3y-5)(z+ 3y-1)) or (x-2y)(2x + 3y-5)(2x+3y-1))

I hope it helps in factoring systematically