Am I doing these right? Help desperatley needed!

RedheadSTL

New member
Joined
Aug 14, 2006
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7
Please tell me if I'm doing this right, and if not, what am I doing wrong. I'm so confused!

Factor Completly:

27X^3 + 9y^3:

(27x-27)(27x^2 + 729X + 729)

64x^3 + 9y^3:

(64X^2 - 675XY + 9Y^2)

Solve for X

6X^2 - X - 15 = 0

3(2x^2 - X-5)

10X^2 - 90 = 0

10(x^2 - 9)
 
RedheadSTL said:
Solve for X
6X^2 - X - 15 = 0
3(2x^2 - X-5)
Red, your attempt is VERY wrong:
this is quite basic stuff; WHY have you got no idea?
No teacher? Learning on your own?
 
You need to solve for x. Therefore, you have to get x by itself.

First, factor 6x^2 - x - 15. When factored 6x^2 - x - 15 becomes

. . .(3x - 5)(2x + 3)

*You can check that this factorization is correct by FOIL-ing it and getting the original polynomial.*

Now get x by itself by making the factors equal to zero (0).

. . .(3x - 5)(2x + 3) = 0

so

. . .(3x - 5) = 0 or (2x + 3) = 0

Start with 3x - 5 = 0. The object is to leave x on one side. Leave it on the left side for this problem. Then subtract 5 to cancel the -5.

. . .3x - 5 = 0
. . .3x - 5 + 5 = 0 + 5
. . .3x = 5

Divide both sides by 3 to get X by itself.

. . .3x/3 = 5/3
. . .x = 5/3

Now do 2x + 3 = 0 the same way as shown above.
 
Thanks so much!

I have been out of school now for 15 years and returning to complete my degree. This is my last class and I'm having such a difficult time with it. I can compose a 20-page paper in 3-hours and ace it, but it takes a lot of time for me to learn and master these concepts.

Everyone has their own cup-of-tea I guess. I wish I was able to understand all of this without extra help. It's so frustrating.
 
You're welcome. Do you mind helping me do a 20-page paper when the time comes? j/k heh. Good luck in your class and getting your degree.
 
So with 10X^2 - 90 = 0

This one is a binomial so it will be completed differently.

5(x^2 - 18) = 0

I'm not sure how I would complete this problem.
 
RedheadSTL said:
So with 10X^2 - 90 = 0

This one is a binomial so it will be completed differently.

5(x^2 - 18) = 0

I'm not sure how I would complete this problem.

Yes, a binomial like this is worked differently. I have to go somewhere right quick. I'll post back witihin 30 minutes if someone else doesn't work it out before me.
 
RedheadSTL said:
So with 10X^2 - 90 = 0
This one is a binomial so it will be completed differently.
5(x^2 - 18) = 0
I'm not sure how I would complete this problem.
Slow down and relax, Red :wink:

First, divide each term by 10; you get:
x^2 - 9 = 0

Now factor that:
(x - 3)(x + 3) = 0

Can you finish it?
 
10X^2 - 90 = 0

10(x^2 - 9)=0
10(x - 3)(x + 3)=0

x-3=0 x+3=0
Solve from there

Something happens with the 10 where you no longer have to worry about it. I don't know why. I just know how to do the problems on sight.
 
Okay! Thank you both. Yes, from here it's pretty easy. :)

I'm terribly sorry, I feel silly not being able to wrap my brain around some of these questions. I only have 3 weeks left in this class!!!
 
letsgetaway said:
10X^2 - 90 = 0

10(x^2 - 9)=0
There is no need to SHOW the 10, lga; it's a simple division by 10;
so x^2 - 9 = 0 is all that's needed.

Same as if you have 10x - 50 = 0; then x - 5 = 0;
you don't go: 10(x - 5) = 0, right?
 
Denis said:
letsgetaway said:
10X^2 - 90 = 0

10(x^2 - 9)=0
There is no need to SHOW the 10, lga; it's a simple division by 10;
so x^2 - 9 = 0 is all that's needed.

Same as if you have 10x - 50 = 0; then x - 5 = 0;
you don't go: 10(x - 5) = 0, right?

true. i just like to show all steps
 
letsgetaway said:
Denis said:
letsgetaway said:
10X^2 - 90 = 0
10(x^2 - 9)=0
There is no need to SHOW the 10, lga; it's a simple division by 10;
so x^2 - 9 = 0 is all that's needed.
Same as if you have 10x - 50 = 0; then x - 5 = 0;
you don't go: 10(x - 5) = 0, right?
true. i just like to show all steps

BUT that's not showing steps; to show steps in dividing 10x^2 - 90 = 0
by 10, you would proceed this way:
10x^2/10 - 90/10 = 0
x^2 - 9 = 0
 
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