text: # challenge question -- Factor the polynomial completely
  
  #### lookagain
  
  ##### Elite Member
  Edit:
  
  Demonstrate at least two methods for factoring the following polynomial
  completely over the integers.
  
  $$\displaystyle x^5 + x^4 + x^3 + x^2 + x + 1$$
  
  Last edited:
  
  #### MarkFL
  
  ##### Super Moderator
  Staff member
  Method 1:
  
  Factor first by grouping:
  
  $$\displaystyle (x^5+x^4)+(x^3+x^2)+(x+1)=x^4(x+1)+x^2(x+1)+(x+1)=(x+1)(x^4+x^2+1)$$
  
  Now, for the quartic factor assume it may be factored as follows:
  
  $$\displaystyle x^4+x^2+1=(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1$$
  
  Equating coefficients, we find:
  
  $$\displaystyle a+b=0$$
  
  $$\displaystyle ab+2=1$$
  
  and so one solution is $$\displaystyle (a,b)=(1,-1)$$ and we have:
  
  $$\displaystyle x^4+x^2+1=(x^2+x+1)(x^2-x+1)$$ which means:
  
  $$\displaystyle x^5+x^4+x^3+x^2+x+1=(x+1)(x^2+x+1)(x^2-x+1)$$
  
  Method 2:
  
  Let:
  
  $$\displaystyle S=x^5+x^4+x^3+x^2+x+1$$ and so:
  
  $$\displaystyle Sx=x^6+x^5+x^4+x^3+x^2+x=S+x^6-1$$ hence:
  
  $$\displaystyle S(x-1)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$$ thus:
  
  $$\displaystyle S=(x+1)(x^2-x+1)(x^2+x+1)$$
  
  #### Subhotosh Khan
  
  ##### Super Moderator
  Staff member
  A) Give the completely factored form over the integers of the following polynomial, and
  
  B) demonstrate at least two methods for doing so.
  
  $$\displaystyle x^5 + x^4 + x^3 + x^2 + x + 1$$
  
  (x+1)(x4+x2+1) → (x+1)(x4+2x2+1 - x2) → (x+1)(x2+1+x)(x2+1-x)
  
  or
  
  (x3+1)(x2+x+1) →(x+1)(x2-x+1)(x2+x+1)
  
  #### lookagain
  
  ##### Elite Member
  Here is another:
  
  (x^5 + 1) + (x^4 + x^3 + x^2 + x) =
  
  (x + 1)(x^4 - x^3 + x^2 - x + 1) + (x^4 + x^3) + (x^2 + x) =
  
  (x + 1)(x^4 - x^3 + x^2 - x + 1) + x^3(x + 1) + x(x + 1) =
  
  (x + 1)(x^4 - x^3 + x^2 - x + 1 + x^3 + x) =
  
  (x + 1)(x^4 + x^2 + 1) =
  
  (You can continue simplifying this with one of the above steps in
  any of the appropriate above posts.)
  
  Last edited:
  
  #### daon2
  
  ##### Senior Member
  Here's another way.
  
  Since $$\displaystyle x^5+x^4+x^3+x^2+x+1 = \dfrac{x^6-1}{x-1}$$ the roots of this polynomial are exactly the set $$\displaystyle \{z\in \mathbb{C}-\{1\}\,\,;\,\, z^6=1\}$$, i.e. the roots of unity, ignoring the positive real root. They are $$\displaystyle e^{\pm i\pi/3},e^{\pm i2\pi/3}, -1$$.
  
  We want a real factorization obviously, and we can see that the conjugate pair to each root is present (as it should be). Pairing them off we get the (minimal) polynomials for each:
  
  $$\displaystyle (x-e^{i\pi/3})(x-e^{- i\pi/3}) = x^2-x+1$$
  $$\displaystyle (x-e^{i2\pi/3})(x-e^{- i2\pi/3}) = x^2+x+1$$
  $$\displaystyle x-(-1) = x+1$$
  
  There is an abstract algebra/number theoretic variation of the above that can be performed for the general case too.
  
  #### soroban
  
  ##### Elite Member
  Hello, lookagain
  
  This is a variation of daon's solution.
  
  Demonstrate at least two methods for factoring the following polynomial:
  
  . . . $$\displaystyle P(x) \;=\;x^5 + x^4 + x^3 + x^2 + x + 1$$
  
  $$\displaystyle P(x) \;=\;\dfrac{x^6 - 1}{x-1} \;=\;\dfrac{\overbrace{(x^3)^2 - (1^2)}^{\text{diff. of squares}}}{x-1}$$
  
  . . . . .$$\displaystyle =\;\dfrac{\overbrace{(x^3-1)}^{\text{diff.of cubes}}\cdot\overbrace{(x^3+1)}^{\text{sum of cubes}}}{x-1}$$
  
  . . . . .$$\displaystyle =\; \dfrac{(\color{red}{\rlap{/////}}{x-1})(x^2+x+1)(x+1)(x^2-x+1)}{\color{red}{\rlap{/////}}x-1}$$
  
  . . . . .$$\displaystyle =\; (x+1)(x^2+x+1)(x^2-x+1)$$
  
  #### lookagain
  
  ##### Elite Member
  Hello, lookagain
  
  This is a variation of daon's solution.
  
  $$\displaystyle P(x) \;=\;\dfrac{x^6 - 1}{x-1} \;=\;\dfrac{\overbrace{(x^3)^2 - (1^2)}^{\text{diff. of squares}}}{x-1}$$
  
  . . . . .$$\displaystyle =\;\dfrac{\overbrace{(x^3-1)}^{\text{diff.of cubes}}\cdot\overbrace{(x^3+1)}^{\text{sum of cubes}}}{x-1}$$
  
  . . . . .$$\displaystyle =\; \dfrac{(\color{red}{\rlap{/////}}{x-1})(x^2+x+1)(x+1)(x^2-x+1)}{\color{red}{\rlap{/////}}x-1}$$
  
  . . . . .$$\displaystyle =\; (x+1)(x^2+x+1)(x^2-x+1)$$
  
  MarkFL said:
  Method 2:
  
  Let:
  
  $$\displaystyle S=x^5+x^4+x^3+x^2+x+1$$ and so:
  
  $$\displaystyle Sx=x^6+x^5+x^4+x^3+x^2+x=S+x^6-1$$ hence:
  
  $$\displaystyle S(x-1)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$$ thus:
  
  $$\displaystyle S=(x+1)(x^2-x+1)(x^2+x+1)$$
  
  These two (MarkFL's and soroban's versions) look essentially the same to me.
  
  - - - - - - - - - - - - - - - - -
  
  Others:
  
  (x^5 + x^2) + (x^4 + x) + (x^3 + 1) =
  
  x^2(x^3 + 1) + x(x^3 + 1) + 1(x^3 + 1) =
  
  (x^3 + 1)(x^2 + x + 1) =
  
  (x + 1)(x^2 - x + 1)(x^2 + x + 1)
  
  . . . . . . . . . . . . . . . . . . . . .
  
  (x^5 + x^3 + x) + (x^4 + x^2 + 1) =
  
  x(x^4 + x^2 + 1) + 1(x^4 + x^2 + 1) =
  
  (x^4 + x^2 + 1)(x + 1) =
  
  (x^2 - x + 1)(x^2 + x + 1)(x + 1)
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