Really Really Urgent

**happyboy** · July 13th 2006, 10:17 AM

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

**galactus** · July 13th 2006, 10:28 AM

For number 2:

It's just a division/cancellation problem.

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

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

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

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

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

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

Now, you finish?.

**happyboy** · July 13th 2006, 10:31 AM

Originally Posted by galactus

For number 2:

It's just a division/cancellation problem.

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

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

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

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

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

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

Now, you finish?.

how would I finish? I'm not trying to act stupid

would it be A - 6 for the answer?

**galactus** · July 13th 2006, 10:39 AM

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:

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

Now cancel what needs cancelled and that's it.

**happyboy** · July 13th 2006, 10:42 AM

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:

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

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

**galactus** · July 13th 2006, 10:43 AM

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.

**happyboy** · July 13th 2006, 10:47 AM

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

**Quick** · July 13th 2006, 11:00 AM

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 $3x^2-2xy+2y^2$ ?

**happyboy** · July 13th 2006, 11:02 AM

Originally Posted by Quick

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

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

**Quick** · July 13th 2006, 11:04 AM

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

Alright, write it out...

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

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

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

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

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

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

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

**happyboy** · July 13th 2006, 11:05 AM

Originally Posted by Quick

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

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

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