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

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

How to solve or factorize this equation?
$\displaystyle x^4 - x^3 + 8x - 8 = 0$

2. 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).

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

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

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

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

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

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

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

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

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

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

5. Originally Posted by wingless
$\displaystyle x^4 - x^3 + 8x - 8 = 0$

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

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

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

$\displaystyle x=\{-2,1\}$
You missed out on two complex solutions.
$\displaystyle x^4 - x^3 + 8x - 8 = 0$

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

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

Now let each factor be zero and solve.

-Dan