Hello, I'am having problems with factoring 40X^3 - 5 can anyone show me how to do this?

- on Sun Mar 13, 2011 10:00 am
- Forum: Beginning Algebra
- Topic: Factoring 40x^3 - 5
- Replies:
**2** - Views:
**1686**

OK, thanks for the relply, I know thanks to the Wikipedia factoring page you send me, how to solve this problem now. I divide each term by 5 to get 8x^3 - 1 that turns in to (2x)^3 - 1^3, then 2x - 1 is a factor, the other factor is found by long division of (2x^3 - 1)/ (2x - 1) = 4x^2 + 2x + 1. I n...

- on Sun Mar 13, 2011 8:23 pm
- Forum: Beginning Algebra
- Topic: Factoring 40x^3 - 5
- Replies:
**2** - Views:
**1686**

Is there an easy way to solve a system like this one?

x y/(3x-4y)=2/11

y z/(2y+3z)=6/5

x z/(x-z)=3/2

Luke

x y/(3x-4y)=2/11

y z/(2y+3z)=6/5

x z/(x-z)=3/2

Luke

- on Thu Apr 07, 2011 7:57 am
- Forum: Beginning Algebra
- Topic: system of linear equations
- Replies:
**10** - Views:
**4051**

Can't solve this one, so please help. A train is travelling from A to B. After it travelled for one hour, it stops for one hour and then continues it's voyage with a speed that is equal of 3/5 of the initial speed. It arrives at B with a delay of 3 hours . If the train would have stopped 50 km furth...

- on Thu Apr 07, 2011 8:22 am
- Forum: Beginning Algebra
- Topic: Word problem: A train is travelling from A to B.
- Replies:
**3** - Views:
**1599**

Giving another example, of how it should be done according to my textbook: given the following system: x y/(x+y)=8/3 y z/(y+z)=8/5 x z/(x+z)=4/3 the first eqn is the same as: (1/x)+(1/y)=3/8 (1) the second eqn is: (1/y)+(1/z)=5/8 (2) the third eqn: (1/x)+(1/z)=6/8 (3) So the sum of the three last eq...

- on Thu Apr 07, 2011 4:48 pm
- Forum: Beginning Algebra
- Topic: system of linear equations
- Replies:
**10** - Views:
**4051**

x^2+7x-144=0

x=9 or x=-16

x=9, since a rectangle has no negative lenght.

9*16=144

x=9 or x=-16

x=9, since a rectangle has no negative lenght.

9*16=144

- on Thu Apr 07, 2011 7:41 pm
- Forum: Beginning Algebra
- Topic: How to solve a stated problem
- Replies:
**1** - Views:
**1142**

Is there an easy way to solve a system like this one? x y/(3x-4y)=2/11 y z/(2y+3z)=6/5 x z/(x-z)=3/2 Luke Can one transform the three equations into the 1/x + 1/y form like in the previous example? Applying the tranformation to the third eqn: x z/(x-z) = 3/2, would be something like: (1/z) - (1/x) =...

- on Fri Apr 08, 2011 8:32 am
- Forum: Beginning Algebra
- Topic: system of linear equations
- Replies:
**10** - Views:
**4051**

distance(ab) = d(ab); v = initial speed; t = normal time of arrival. d(ab) = v * t first case: d(ab) = v *1 + 3/5 *v *(t+3 -1) d(ab) = v + 3/5 v * (t+2) d(ab) = v + 3/5 * v *t + 6/5 * v second case: d(ab) = v + 50 + 3/5 *v * (t +3 -1 - 1.66) d(ab)= v + 50 +3/5 v *t + 1/5 * v since the distance is th...

- on Fri Apr 08, 2011 2:52 pm
- Forum: Beginning Algebra
- Topic: Word problem: A train is travelling from A to B.
- Replies:
**3** - Views:
**1599**

I don't know how to do this so please help.

Luke.

Luke.

- on Fri Apr 08, 2011 3:03 pm
- Forum: Beginning Algebra
- Topic: system of linear equations
- Replies:
**10** - Views:
**4051**

I know that : x y /(x+y) = 8/3 is the same as 1/x + 1/y = 3/8

Could this sort of transformation also be done with: x y / (3x - 4y) = 2/11 ? ( guess not, but I'm not sure).

Thanks.

Could this sort of transformation also be done with: x y / (3x - 4y) = 2/11 ? ( guess not, but I'm not sure).

Thanks.

- on Sat Apr 09, 2011 1:34 pm
- Forum: Beginning Algebra
- Topic: system of linear equations
- Replies:
**10** - Views:
**4051**