How to solve or factorize this equation?

$\displaystyle x^4 - x^3 + 8x - 8 = 0$

help please

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- Mar 16th 2008, 06:49 AMzxeSolve x^4 - x^3 + 8x - 8 = 0, help pls
How to solve or factorize this equation?

$\displaystyle x^4 - x^3 + 8x - 8 = 0$

help please - Mar 16th 2008, 06:55 AMMoo
Hello,

You can observe that the sum of the coefficients is 0. So 1 is solution. This means that you can factorise by (x-1).

So $\displaystyle x^4-x^3+8x-8 = (x-1)Q(x)$

Where Q(x) is a polynom of degree 3 -> Q(x)=ax^3 + bx² + cx + d, with a,b,c,d to determine by developping (x-1)Q(x). - Mar 16th 2008, 06:55 AMwingless
$\displaystyle x^4 - x^3 + 8x - 8 = 0$

$\displaystyle x^3(x-1) + 8(x-1) = 0$

$\displaystyle (x-1)(x^3+8) = 0$

---> $\displaystyle x-1 = 0$

---> $\displaystyle x^3+8 = 0$

$\displaystyle x=\{-2,1\}$ - Mar 16th 2008, 08:09 AMJhevon
wingless is right. i'd just like to point out that we can solve $\displaystyle x^2 + 8 = 0$ using the sum of two cubes formula (if you don't see the solution immediately). so, $\displaystyle x^3 + 8 = (x + 2)(x^2 - 2x + 2^2) = 0$, and now equate each term in the product to zero and solve.

- Mar 16th 2008, 12:28 PMtopsquark