# Solve x^4 - x^3 + 8x - 8 = 0, help pls

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

• Mar 16th 2008, 06:55 AM
Moo
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 $x^4-x^3+8x-8 = (x-1)Q(x)$

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 AM
wingless
$x^4 - x^3 + 8x - 8 = 0$

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

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

---> $x-1 = 0$
---> $x^3+8 = 0$

$x=\{-2,1\}$
• Mar 16th 2008, 08:09 AM
Jhevon
Quote:

Originally Posted by wingless
$x^4 - x^3 + 8x - 8 = 0$

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

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

---> $x-1 = 0$
---> $x^3+8 = 0$

$x=\{-2,1\}$

wingless is right. i'd just like to point out that we can solve $x^2 + 8 = 0$ using the sum of two cubes formula (if you don't see the solution immediately). so, $x^3 + 8 = (x + 2)(x^2 - 2x + 2^2) = 0$, and now equate each term in the product to zero and solve.
• Mar 16th 2008, 12:28 PM
topsquark
Quote:

Originally Posted by wingless
$x^4 - x^3 + 8x - 8 = 0$

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

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

---> $x-1 = 0$
---> $x^3+8 = 0$

$x=\{-2,1\}$

You missed out on two complex solutions.
$x^4 - x^3 + 8x - 8 = 0$

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

$(x - 1)(x + 2)(x^2 - 2x + 4) = 0$

Now let each factor be zero and solve.

-Dan