Thread: Really Really Urgent

I REALLY NEED HELP WITH THESE TWO PROBLEMS, please someone help solveeeeee i'm begginggg you, I keep getting them wrong everytime I do them

How would I multiply this problem? the small numbers are twos
http://img397.imageshack.us/img397/7599/math13ep.jpg



How would I perform the indicated operation on this problem?
http://img397.imageshack.us/img397/9062/math20me.jpg

For number 2:

It's just a division/cancellation problem.

$\displaystyle \frac{a^{2}-a-30}{a+5}$

Factor the numerator and you will quickly see the cancellations.

If you're having a rough time factoring, ask yourself, what 2 numbers when multiplied equal -1 and when added equal -30

How about -6 and 5

(-6)(5)=-30 and -6+5=-1

$\displaystyle a^{2}-6a+5a-30$

$\displaystyle (a^{2}-6a)+(5a-30)$

$\displaystyle a(a-6)+5(a-6)$

$\displaystyle (a+5)(a-6)$

Now, you finish?.

Originally Posted by galactus

For number 2:

It's just a division/cancellation problem.

$\displaystyle \frac{a^{2}-a-30}{a+5}$

Factor the numerator and you will quickly see the cancellations.

If you're having a rough time factoring, ask yourself, what 2 numbers when multiplied equal -1 and when added equal -30

How about -6 and 5

(-6)(5)=-30 and -6+5=-1

$\displaystyle a^{2}-6a+5a-30$

$\displaystyle (a^{2}-6a)+(5a-30)$

$\displaystyle a(a-6)+5(a-6)$

$\displaystyle (a+5)(a-6)$

Now, you finish?.

how would I finish? I'm not trying to act stupid
would it be A - 6 for the answer?

You ought to go see your teacher if you're still lost from this point on.

Just some friendly advice.

Here's all is left:

$\displaystyle \frac{(a-6)(a+5)}{a+5}$

Now cancel what needs cancelled and that's it.

Originally Posted by galactus

You ought to go see your teacher if you're still lost from this point on.

Just some friendly advice.

Here's all is left:

$\displaystyle \frac{(a-6)(a+5)}{a+5}$

Now cancel what needs cancelled and that's it.

Its homeschool and its self-taught, I'm going into a normal school next year, I Just need these two problems and im finished with this grade

Originally Posted by happyboy

how would I finish? I'm not trying to act stupid
would it be A - 6 for the answer?

Yes, you got it! . You updated your post while I was writing mine.

Anytime you see a division problem set up like the one you have, rewrite it the way I did and see whether or not anything factors. If it does, they'll more than likely be some cancellations and you'll be on your way.

Originally Posted by galactus

Yes, you got it! . You updated your post while I was writing mine.

Anytime you see a division problem set up like the one you have, rewrite it the way I did and see whether or not anything factors. If it does, they'll more than likely be some cancellations and you'll be on your way.

I just have one more problem and i'm done, I don't get how to do the other one at all tho

Originally Posted by happyboy

I just have one more problem and i'm done, I don't get how to do the other one at all tho

I don't understand question #1 do you want us to factor $\displaystyle 3x^2-2xy+2y^2$?

Originally Posted by Quick

I don't understand question #1 do you want us to factor $\displaystyle 3x^2-2xy+2y^2$?

I don't understand it either, the x - y under it is apart of the problem too. and all it says is to multiply

Originally Posted by happyboy

I don't understand it either, the x - y under it is apart of the problem too. and all it says is to multiply

So you want us to do $\displaystyle \left(3x^2-2xy+2y^2\right)\times\left(x-y\right)$?

Alright, write it out...

$\displaystyle \left(x-y\right)\left(3x^2-2xy+2y^2\right)$

use the distributive property:
$\displaystyle \left(3x^2\left(x-y\right)-2xy\left(x-y\right)+2y^2\left(x-y\right)\right)$

use the distibutive property again: $\displaystyle 3x^2\left(x\right)-3x^2\left(y\right)-2xy\left(x\right)+2xy\left(y\right)+2y^2\left(x
\right)-2y^2\left(y\right)$

multiply: $\displaystyle 3x^3-3x^2y-2x^2y+2y^2x+2y^2x-2y^3$

group like terms:$\displaystyle 3x^3-(3x^2y+2x^2y)+(2y^2x+2y^2x)-2y^3$

add/subtract: $\displaystyle \boxed{3x^3-5x^2y+4y^2x-2y^3}$

I think that's the answer you want (if I understand the question correctly)

Originally Posted by Quick

So you want us to do $\displaystyle \left(3x^2-2xy+2y^2\right)\times\left(x-y\right)$?

I'm guessing that is it, all the problem said was multiply and x-y was under it. this is a really confusing problem