I figured since I don't want to clutter up this board with a bunch of newly created threads, that I'd post whatever math problems I'm having in here.

I suck at math, sadly.

 

The most recent problem I've come across is this:

 

factor each expression completely

 

x^3 - 8y^3

 

So I tried doing this since: (x-y)(x^2 + xy + y^2)

 

factored:

 

x^3 + 2x^2y + 4xy^2 - 2x^2y - 4xy^2 - 8y^3

 

Which turns out to be:

 

(x -2y)(x + 2xy + 4y)

 

what I don't understand from this problem is how 4 gets there.

Try looking at the formula for the difference of cubes.

x^3 – y^3 = (x – y) (x^2 + xy +y^2)

x^3 – 8y^3 is of the form a^3 – b^3, with a=x, b=2y. Also b^3 comes out of the factors by multiplying b*b^2. And, in this case, b=2y and b^2 = 4y^2. So, that's where the 4 comes from.

 

PS : There's a typo in your final factorization. The last term is 4y^2, not 4y.

The 4 part is what is getting me.

From what I figured:

 

[math] (x - 2y)(x^2 + 2xy + 2y^2) [/math]

 

what I don't get here is how there is suppose to be a [math]4[/math] in place of the [math]2[/math] in [math]2y^2.[/math]

 

If i factor out 8

 

[math]2 * 2 * 2 = 8[/math]

 

I don't understand the math reasoning to put a [math]4[/math] there.

8 = 2*4

 

Multiply out (x - 2y)(x^2 + 2xy + 2y^2) and see if you get x^3 - 8y^3. You will not ! Simply because (-2y)(2y^2) = -4y^3 instead of -8y^3, which is = (-2y)(4y^2)

I figured out that there is an x, y, and z variable in this word problem, but figuring it all out is somewhat confusing for me.

 

Word Problem:

 

The sum of three integers is 18. The third integer is four times the second, and the second integer is 6 more than the first. Find the integers.

 

(x, y, z)

well start with x+y+z=18, z=4y, and y=6+x. The 3rd can be changed to y-6=x, then you can substitute the two back into the first to get (y-6)+y+4y=18, solve for y and just plug them back into the other equations to get the rest

Here are a few problems I've come across. I think I understand Gaussian Elimination but it just confuses me to some extents.

 

Instructions: Use matrices to solve each system of equations. Each system has one solution.

 

I'm i'm looking at:

 

2x + y + 3z = 3

-2x - y + z = 5

4x - 2y - 3z = 10

 

So I decided to turn it into

 

[ 2 1 3|3]

[-2 -1 1|5]

[4 -2 2|2]

 

So from there I decided that when doing this gaussian thing the goal is to eliminate all numbers below the first digit in the first row to 0.

 

I decided that I didn't need to exchange rows.

 

R=Row

R1 + R2 = [0 0 4 | 8 ] (new R2)

 

Since that was done with I decided that I now had to move to row three.

 

So to get rid of the 4 I have to make the first row negative.

 

R1 * -2 = [ -4 -2 -6 |-3 ]

 

[ -4 -2 -6 |-3 ]

+

[ 4 -2 2|2]

----------------------

[0 -4 -4 |-1]

 

I don't really know if I did this right, and I don't know where to go after this step.

You want to get the matrix in this form :

1 0 0 | x

0 1 0 | y

0 0 1 | z

 

The new R2 you did qualifies for the R3 of above if you divide it with 4.

Turns out that z = 2. So we got R3 =[ 0 0 1 | 2].

The new R1 you did (slight mistake, u forgot to multiply smth there[-3 should be -6], anyway, it is 0 4 4 | 4 if multiplied with -1. This is good for our R2.

R2 = R2 - 4*R3 = 0 4 0 | -4 and looks like that y = -1. R2 =[ 0 1 0 | -1].

Now, your original R1 [2 1 3 | 3] can be turned R1 = R1 - R2 - 3*R3 = [2 0 0 | -2 ] -> R1 = [1 0 0 | -1] and x is also -1.

 

I hope that wasn't too confusing, it's been a while I did this

4x - 2y - 3z = 10

[4 -2 2|2]

If the original problem is from your book, your answer from this approach won't be equivalent.

I think you put the no. wrongly into matrix. This is the major problem I suppose.