# need help factoring a problem

• November 5th 2010, 07:37 PM
NecroWinter
need help factoring a problem
9x^2y^2+5xy-4

I dont really even know where to begin (Wondering)

My book doesnt explain what to do with this kind of problem at all. I mean it does talk about factoring trinomials, but when its setup this way its a bit confusing, and different

any help is appreciated.
• November 5th 2010, 07:42 PM
pickslides
It is a tricky one.

Maybe before trying this one, have a go at $9a^2+5a-4$

Can you do it?
• November 6th 2010, 08:47 AM
NecroWinter
yeah I would rewrite the problem as

(9a2 + 9a)(-4a - 4)
9a(a + 1) -4(a+1)
(9a-4)(a+1)
• November 6th 2010, 09:18 AM
Wilmer
Good; do same with yours, after letting y=1 ; get it?
• November 6th 2010, 10:01 AM
NecroWinter
Quote:

Originally Posted by Wilmer
Good; do same with yours, after letting y=1 ; get it?

unfortunately im a little confused by what you mean lol
• November 6th 2010, 10:34 AM
skeeter
Quote:

Originally Posted by NecroWinter
unfortunately im a little confused by what you mean lol

$9a^2 + 5a - 4$

$9(xy)^2 + 5(xy) - 4$
• November 6th 2010, 11:40 AM
NecroWinter
Ok so heres what I think I should do with that, if someone can fix my train of thought i think that would help me correct the errors.

9x^2y^2+9x-4x-4
(9x^2y^2+9x)(-4x-4)

(9x^2y^2+9x)
9x(y^2 + 1)-4(x+1)

this would be so much better if my book actually talked about factoring trinomials when you have exponents set up this way, outside the norm

i also tried something like this but it just didnt look right at all:
9(xy)^2+5(xy)-4

9xy(1^2 +1) -4(1(xy) +1)
• November 6th 2010, 11:52 AM
skeeter
you're really making this much more difficult than it needs to be ...

pickslides had you factor $9a^2 + 5a - 4$ for a reason.

$9a^2 + 5a - 4 = (9a-4)(a+1)$

compare the factorization with $a$ above to the factorization with $xy$ below ...

$9(xy)^2 + 5(xy) - 4 = (9xy - 4)(xy + 1)$

see the similarity?
• November 6th 2010, 12:30 PM
NecroWinter
Okay I gotcha. Thanks

Is this typically how you guys tackle these kinds of problems? Will simply substituting(for a lack of a better word) make me run into a problem later on?
• November 6th 2010, 12:46 PM
skeeter
can't say ... depends on the problem.