/sci/ - Science & Math

>>1146953
>I could just "guess"

So do it motherfucker.

>>1146967
there sure is another method

Cubic equations have a solving formula.

There is a way to factor it, but I don't know what fancy names you foreigners give it.

check algorithm for solving equation of 3rd grade on wiki.

>>1146970

>>1146953

search "depressed cubic" in google. this is a cubic equation without an x^2 term. every cubic can be transformed into one of these and every depressed cubic can then be solved algebraically by a few methods, cardano's formula

the only problem is that sometime's the answer may seem a bit convoluted. the answer 2 may come out as the sum of 2 cube roots that arent directly identifiable as adding to an integer value

it can't be helped. we'll have to use "THAT"

x^3 + x - 10 = 0
2 goes into it

x^3 + 0x^2 + x - 10 / (x-2)
whatever you get is the other root

ok thats the case with cubic ones, thank you, but what do if I have an equation like in this picture?

its derivative is x^2+1 > 0
thus the function is increasing
thus it has at most one root
2 is a root
yupi

>>1147012
polynomial division, same as cubic.

>>1147012
you can do what this awesome person >>1147024 suggested

>>1147012
1024 is a perfect 5th power
answer is 4

>>1147054
>1024
>1032

>>1147012

PROTIP : 2^10 = 1024

i would use newton's method

<div class="math">\usepackage{amssymb}\begin{eqnarray*} r_1 & = & -\frac{a}{3} + \left(\frac{-2a^3 + 9ab - 27c + \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\ & & {} + \left(\frac{-2a^3 + 9ab - 27c - \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\ r_2 & = & -\frac{a}{3} - \frac{1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c + \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\ & & {} + \frac{-1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c - \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\ r_3 & = & -\frac{a}{3} + \frac{-1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c + \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\ & & {} - \frac{1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c - \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \end{eqnarray*}</div>

>>1147060

4^5 +2(4) = 1024 + 8 = 1032

>>1147060
4^5+2*4

>>1147072
i guess i should explain this

your function is x^3 + x - 10 = 0

so your derivative is 3x^2 + 1

newton's method is basically just guessing a number that you think it is, and then applying the formula

(your new guess) = (your guess) - (f(your guess)/f'(your guess))

and you do that until your new guess is not a new number

god damnit I feel like I'm retarted...
feels bad not having english as native language and discussing maths with it damn

never heard of polynomal division

but thanks for help guys...

Here's a cheap trick.

x^3 + x - 10 = 0
x^3 - 2x^2 + 2x^2 + x -10 = 0 (just added 0 to both sides)

Now factor by grouping,

(x^3 - 2x^2) + (2x^2 + x - 10) = 0
(x^2)(x- 2) + (2x + 5)(x - 2) = 0
(x - 2) (x^2 + 2x + 5) = 0

So x = 2 is a root.