Primenumbers

[Lillen]

To disturb the order a tiny bit... I don't know?!!

 

[Lillen]

May I ask how do i solve x^3 (or higher power then two) equations and fuctions? How do i operate when there are more then two factors. I forgot and cant find the answer to this question on the page factorization on wikipedia, For example (x+3)(x+5)(x+7). I can guess, but i am not certain of the guess!! So how do i operate here?

And i ask this beside the discussion of primes!!

 

[Wiccan_Child]

May I ask how do i solve x^3 (or higher power then two) equations and fuctions? How do i operate when there are more then two factors. I forgot and cant find the answer to this question on the page factorization on wikipedia, For example (x+3)(x+5)(x+7). I can guess, but i am not certain of the guess!! So how do i operate here?

And i ask this beside the discussion of primes!!

The solutions to the equation y = (x + a)(x + b)(x + c) are x = -a, x = -b, and x = -c. So, the solutions to your example would be -3, -5, and -7.

 

[Elendur]

The solutions to the equation y = (x + a)(x + b)(x + c) are x = -a, x = -b, and x = -c. So, the solutions to your example would be -3, -5, and -7.

Corrected that for you

 

[Wiccan_Child]

Corrected that for you

If I was wearing my pedant pendant, technically I was already correct... but I'm not, and I wasn't, so thank you

 

[Elendur]

If I was wearing my pedant pendant, technically I was already correct... but I'm not, and I wasn't, so thank you

no problems.

 

[Lillen]

But that was not the question, i asked how do i operate algebraic (???). This really is junior high math. But i forgot about it already. What you do was using factorization method i think (???), no!?

(x+2)(x+5) you first take the first x and multiply it with the x in the other having x^2 then you take the same x of the first and multiply it with 5 from the second factor, and finally you do the same thing with the '2' in the first factor. How do i do operate when there are three destinct factors???

You see i forgot most about junior high, because in highschool I was always high.

 

[Lillen]

Don't i set x to 0 when i use the factorization method?

 

[Wiccan_Child]

But that was not the question, i asked how do i operate algebraic (???). This really is junior high math. But i forgot about it already. What you do was using factorization method i think (???), no!?

(x+2)(x+5) you first take the first x and multiply it with the x in the other having x^2 then you take the same x of the first and multiply it with 5 from the second factor, and finally you do the same thing with the '2' in the first factor. How do i do operate when there are three destinct factors???

If you're asking how to expand the brackets (there's something of a language barrier here), then you expand the first two brackets, then the resulting terms with the third bracket:

(x + a)(x + b)(x + c)

Multiply each term in the first bracket with each term in the second bracket:

(x[sup]2[/sup] + ax + bx + ab)(x + c)

Repeat (colours show you where the terms come from):

x[sup]3[/sup] + cx[sup]2[/sup] + ax[sup]2[/sup] + acx + bx[sup]2[/sup] + bcx + abx + abc

Simplifying:

x[sup]3[/sup] + (a+b+c)x[sup]2[/sup] + (ab+ac+bc)x + abc

Does that help?

 

[Lillen]

Yeah, that helped!! and i been taught the swedish samantics, not the english, hence the language barrio!!! In swedish everything within the two first standard operational signs (+ and -) is called a term and everything within "()" is called factor if i recall correctly.

 

[Lillen]

Are there other ways?

 

[Lillen]

Let's look at the primes, if we define primes to be the value of which order they come in, then we see that 13/3 = 3 as you take the 6th prime devided with the 2th! and 6 / 2 is 3... the other way is like Wiccan Child promoted - change the definition of the devision operation. And we knew that because we were so bloody special everyone thought we were left behind.

 

[Wiccan_Child]

Yeah, that helped!! and i been taught the swedish samantics, not the english, hence the language barrio!!! In swedish everything within the two first standard operational signs (+ and -) is called a term and everything within "()" is called factor if i recall correctly.

I think that's what they are in English, too. The process I described is 'expanding the brackets', FYI.

 

[Lillen]

Oh...

 

[Lillen]

What if we invent a new way to operate with primes, somehow that will be technically more easy then using my method?!