/sci/ - Science & Math

[Anonymous No.7377022]

So.. I'm retarded and I forgot how to factor this out. It's been a while since I've done math and I'm doing a self-diagonsis test before I start on Calculus again.

Anybody mind doing it for me showing work?

I get as far as to x(x^3+27x)

I can get the answer.. but I have no clue how I got it. I don't want to do math by intuition anymore but truly understand it.

[Anonymous No.7377029]

>>7377022
what are you trying to factor to? just simplify?

also your factor is wrong, you mean this: x(x^3+27)
multiply it out and you'll see

algebra will fuck you in the ass with calc if you're not already confident about it

[Anonymous No.7377032]

Firstly if you're not trolling, so far what you got is wrong. X(x^3+27x) doesn't = x^4 +27x which should be obvious. To figure that one out see if you can distribute the the X in x(x^3+27x) and get back to x^4 + 27x. Secondly, even though you factored it wrong it should be obvious to you that it's looking for something having to do with cubic,(cubic roots maybe?) and you have a cube from the x^3 and if you tried iterating the cubes from 1 to 4 you'd find that 3^3 = 27. Lastly, that's the most help you will get from me and look for the answer in the question. I know it seems redundant for me to day that but you it's pretty obvious that it's a factoring relating to cubes question.

[Anonymous No.7377034]

>>7377029
Oh my bad. That's what I have on my paper. I just couldn't see what I was typing into the chatbox.

I'm trying to factor it into this.

As for my algebraic foundation. It's more solid than the average person, but for some odd reason, this is the one problem in the whole test that I couldn't wrap my head around.

[Anonymous No.7377037]

>>7377034
Fuck. Wrong Answer.

[Anonymous No.7377084]

>>7377037
Simplify it to x^3+27
Of the form x^3+d
The answer is (x+3)(x^2-3x+9)
Of the form (x+b)(x^2-bx+c)
It only works when b^2=c and c*b=d, and b is a real integer
If these conditions aren't settled its a whole lot messier (ie if d isn't a product of a number and its square)

So, just remember the formula and you should be set.

[Anonymous No.7377089]

>>7377037
>>7377034
This is the technique you need: http://www.purplemath.com/modules/polydiv2.htm

The procedure works like this:
- First, by some unknown magic, you somehow deduce that x=0 is a zero of x^4 + 27x. That is, substituting x=0 in that expression yields zero.
- From this, you can conclude that factoring x^4 + 27x will yield, among others, a factor x; that is, the end result of factoring will look like x * (something).
- Using the polynomial long division method linked above, you compute (x^4 + 27x) / (x). This yields x^3 + 27. You conclude that (x^4 + 27x) = (x)(x^3 + 27).
- That's one part of factoring, but you aren't done. The next step is to repeat the factoring procedure for x^3 + 27.
- Repeating this procedure, you magically (that is, using techniques not covered in this post) notice that x=-3 is a zero of x^3 + 27; that is, substituting x=-3 in that yields 0.
- You conclude that (x - -3), AKA (x + 3), must be a factor in the end result.
- Using polynomial long division again, you compute (x^3 + 27) / (x + 3). This yields (x^2 - 3x + 9). Thus, you have (x^3 + 27) = (x + 3)(x^2 - 3x + 9).
- So far, you have factored (x^4 + 27x) = (x)9x + 3)(x^2 - 3x + 9).
- You magically notice that x^2 - 3x + 9 doesn't have any zeroes -- it's positive for any value of x. That means it can't be factored any further, and you're done. (Technically it can be factored into terms containing imaginary numbers, but that's probably beyond the scope of what you're supposed to know.)

So how do you magically know the zeroes x=0 and x=-3? Well, the x=0 is obvious because all the terms in x^4 + 27x have a factor x in them, which makes x=0 a zero. As for x=-3, the hint here is that 27 = 3^3, which means you should check 3 and -3 for being zeroes; that way, you find x=-3 as a zero. In general though, finding zeroes is hard.

While performing a polynomial long division, mind your minus signs. I always fuck them up big time.

[Anonymous No.7377102]

>>7377084
Herp derp, I'm a retard. It actually works for any d, b and c are just the cube root and square of the cube root

[Anonymous No.7377109]

>>7377089
>>7377084
>>7377102

Thanks everybody. It really helped. By far the most helpful board.

[Anonymous No.7377974]

>>7377089
one of the best explanations I have ever read

[Anonymous No.7378016]

>>7377974
what are you some kind of retard? that explanation is for laymen, it wasn't even extensive.. please gtfo of /sci/ and go back to reddit where you feel superior to others but can't do babby fucking polynomial factoring

[Anonymous No.7378019]

>>7378016
>think hes better than me and assumes superiority because I lauded someone for giving a concise and well explained answer

absolutely pathetic

[Anonymous No.7378341]

>>7378019
You lauded your own post, though.

I know because nobody on /sci/ is going to congratulate someone on such a simple explanation.

[Anonymous No.7379645]

>>7378016
factor x^5+x+1.

[Hustler No.7379705]

>>7378341
Why are you such a dick?

Were you picked last today in gym class?

Not everyone has the same strengths as you; some people come from backgrounds full of adversity that force them to focus on survival rather than manipulating numbers on paper.

Clearly OP is trying to learn. You should applaud his efforts of educating himself instead of bashing him.

Seriously go get yourself checked out man.

[Anonymous No.7379850]

>>7377022
Look up 'sum of cubes' and 'difference of cubes' and memorize those formulas tbh.

[Anonymous No.7381413]

>>7377022
the roots are x=0, -3, (3/2)(1+sqrt(-3)) and (3/2)(1-sqrt(-3))

the first root comes from the x factoring out as you observed. then note that x^3+27 has an obvious zero of -3, since cubing it is -27. so you know you can factor out (x+3). use long division to get x^2-3x+9). you can now use good old quadratic formula to get the remaining two roots.

the polynomial has no repeated roots

[Anonymous No.7381429]

this will hep if you read it
>http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.32.6357

[Anonymous No.7381432]

>>7377974
>>7377089
samefag pls go

also american maths is stupid

[Anonymous No.7381478]

>>7377022
<span class="math">x^4+27x=x(x^3+27)[/spoiler]
<span class="math">x^3+27=0\Rightarrow x=-3[/spoiler] by factor theorem
<span class="math">\frac{x^3+27}{x+3}=x^3-3x+9[/spoiler] so
<span class="math">x(x^3+27)=x(x+3)(x^2-3x+9)[/spoiler]

[Anonymous No.7381538]

>>7379705
> Clearly OP is trying to learn
Or get us to do his homework. Which do you really think it is?

[Anonymous No.7381663]

>>7381478
x4+27x=x(x3+27)
x3+27=0x=−3 theorem by factory
x+3x3+27=x3−3x+9 so
x(x3+27)=x(x+3)(x2−3x+9)

[Anonymous No.7382241]

>>7377022
x^4 + 27x
= x(x^3 + 27)
=x(x + 3)(x + 3ω)(x + 3ω^2)