1. ## Factorising

Express x^4 - x in its fully factorise form.

I went through factorising methods:

common factor
difference of two squares X
breaking brackets X
polynomial ?

I would say answer = x(x^3 - 1)
But that is way to simple for the level i'm taking, so im asking if polynomials are involved here? If not, what else should i try?
Thanks

2. Originally Posted by r_maths
Express x^4 - x in its fully factorise form.

I went through factorising methods:

common factor
difference of two squares X
breaking brackets X
polynomial ?

I would say answer = x(x^3 - 1)
But that is way to simple for the level i'm taking, so im asking if polynomials are involved here? If not, what else should i try?
Thanks
You can do more.
$\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$
Thus,
$\displaystyle x(x^3-1)=x(x-1)(x^2+x+1)$

3. Originally Posted by ThePerfectHacker
You can do more.
$\displaystyle x(x^3-1)=x(x-1)(x^2+x+1)$

$\displaystyle (x^2+x+1)$
that could also factorise more?

$\displaystyle (x+1) (x+1)$
?

4. Originally Posted by r_maths
$\displaystyle (x^2+x+1)$
that could also factorise more?

$\displaystyle (x+1) (x+1)$
?
No. (x+1)(x+1)=(x+1)^2=x^2 +2x + 1 =/= x^2 + x +1
if you like to play with that you can always write
((x+1)^2) - x
but I would leave it the way it is x^2 + x +1

5. Originally Posted by Aglaia
No. (x+1)(x+1)=(x+1)^2=x^2 +2x + 1 =/= x^2 + x +1
if you like to play with that you can always write
((x+1)^2) - x
but I would leave it the way it is x^2 + x +1
cheers, i should of checkd with FOIL