Factoring problem

[Ravingsofthesane]

Hello everyone ,

got a issue, before I can fully understand some of the more advanced aspects of factoring I need to get some understanding of this.

here is a basic problem.

9X^2y^2 + 24XY + 16

For this I will use A + B + C for comparison

The only way that I know to complete this problem is to take

A*C which equals 144X^2Y^2

then find the multiple of this result whos product = B

then rewite this problem and factor by grouping.



however the reason for this post is i belive there is another way to solve this without having to go through the hastle of finding all the multiples of 144 whos product = 24 ..


I know that the first part is (3xy ) (3xy )

but what is the next step in this without having to go through all the other steps?


thanks in advance for any help that anyone can offer.

 

[Unco]

however the reason for this post is i belive there is another way to solve this without having to go through the hastle of finding all the multiples of 144 whos sum = 24 ..

...um... does 12 ring a bell?

 

[Ravingsofthesane]

...um... does 12 ring a bell?

well yes , I do realise that 12 is the correct anwser.

but the only way I know to continue from that point is to rewrite and factor by grouping.

The reason for the post was to see if there was another way other then that to solve this.

 

[Unco]

The only "improvement" would be to let t = xy to make things slightly simpler.

You can always just use the quadratic formula.

 

[Denis]

9X^2y^2 + 24XY + 16 = 0

What's sqrt(b^2 - 4ac)? sqrt(576-576) = 0, right?

So that tells us: xy = -b/2a = -24/18 = -4/3

So that "forces" the factoring to be (3xy + 4)(3xy + 4) = 0

Hmmm...Unco ?!

 

[Ravingsofthesane]

ok let me post up a new issue then

here is the problem

-8R^2 + 26R - 15

With this problem I get the following :

-8R^2 + 20R + 6R - 15

when I factor by grouping I get this

-4R(2R - 5) 3(2R-5)

so my anwser would be :

(-4R+3) (2R-5)


checking my anwser i see the correct anwser should be -(4R-3)(2R-5)
can anyone give me a little insight to why this is?

 

[Unco]

Yes, he/she was asking for a better way for the method used above. I suggested the quadratic formula.

Edit: above as in first post.

 

[Unco]

(-4R + 3) = -1(-1(-4R + 3)) = -1(4R - 3) = -(4R - 3)

 

[soroban]

Hello, Ravings!


With a problem like: .\(\displaystyle 9x^2x^2\, +\, 24xy +\, 16\)

Note that the first and last terms are <u>squares</u>.

. . There's a good chance that the trinomial is the square of a binomial.


I would first try: . \(\displaystyle (3xy\;\;\;\underbrace{4)(3xy}\;\;\;4)\)
. . . . . . . . . . . . . . .\(\displaystyle \underbrace{\;\;\;\;\;\;12xy\;\;\;\;\;\;}\)
. . . . . . . . . . . . . . . . . . . . .\(\displaystyle 12xy\)

So the answer is: .\(\displaystyle (3xy\, +\, 4)(3x\, +\, 4) \:= \3xy\, +\, 4)^2\)


Of course, they can always mislead us with, say: .\(\displaystyle x^2\, -\, 13x\, +\, 36\)

. . which is not \(\displaystyle (x\, -\, 6)^2\) . . . but \(\displaystyle (x\, -\, 4)(x\, -\, 9)\)

 

[soroban]

Hello, Moderators!

Lately, there's been a terrible glitch in the LaTeX . . .

I type:
   [tex]-8R^2 + 26R - 15 = -(8R^2 - 26R + 15)[/tex]

And the Preview displays:
\(\displaystyle -8R^2 + 26R - 15 \;= \;-(8R^2 - 26R + 15)\)

Click on QUOTE to see what I actually typed.


Previously, I could retype the entire line.
Now it seems that I'm stuck with the \(\displaystyle f(x) = x^2\)
. . no matter what I type.