1. ## Factorisation

(1 + x + x^2 + x^3)^2 - x^3

2. Hello anshulbshah
Originally Posted by anshulbshah
(1 + x + x^2 + x^3)^2 - x^3
See the attached graph, which shows that $f(x)=(1 + x + x^2 + x^3)^2 - x^3 =0$ doesn't have any real roots, and hence $(1 + x + x^2 + x^3)^2 - x^3$ doesn't have any factors.

$f(-1)=f(0)=1$, and there's a minimum turning point between these two values, but this minimum value is greater than zero. You could possibly prove this by calculus.

For $x>0$ and $x<-1$, $f(x)$ increases very rapidly, since the dominant term when the brackets are expanded is $x^6$.

3. it can be solve by grouping after expanding,

(1 + x + x^2 + x^3)^2 - x^3 = (x^2 + x + 1)(x^4 + x^3 + x^2 + x + 1)

4. Originally Posted by pacman
it can be solve by grouping after expanding,

(1 + x + x^2 + x^3)^2 - x^3 = (x^2 + x + 1)(x^4 + x^3 + x^2 + x + 1)
Thanks for this - I was thinking of a linear factor, of course!

5. Prove: (1 + x + x^2 + x^3)^2 - x^3 = (x^2 + x + 1)(x^4 + x^3 + x^2 + x + 1)

[(1 + x) + (x^2 + x^3)]^2 - x^3 = (1 + x)^2 + 2(1 + x)(x^2 + x^3) + (x^2 + x^3)^2 - x^3,

= (1 + 2x + x^2) + (2x^2 + 2x^3 +2x^3 + 2x^4) + (x^4 + 2x^5 + x^6) - x^3, rearranging

= (1) + (2x) + (x^2 + 2x^2) + (2x^3 +2x^3 - x^3) + (2x^4 + x^4) + (2x^5) + (x^6),

= 1 + 2x + 3x^2 + 3x^3 + 3x^4 + 2x^5 + x^6, sub-divide with numerical coefficient equal to 1, group them THEN apply factoring

= (1 + x + x^2) + (x + x^2 + x^3) + (x^2 + x^3 + x^4) + (x^3 + x^4 + x^5) + (x^4 + x^5 + x^6)

= (1 + x + x^2) + x(1 + x + x^2) + x^2(1 + x + x^2) + x^3(1 + x + x^2) + x^4(1 + x + x^2), factoring

= (1 + x + x^2)(1 + x + x^2 + x^3 + x^4).