# Thread: challenge question -- Factor the polynomial completely

1. ## challenge question -- Factor the polynomial completely

Edit:

Demonstrate at least two methods for factoring the following polynomial
completely over the integers.

$x^5 + x^4 + x^3 + x^2 + x + 1$

2. Method 1:

Factor first by grouping:

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

Now, for the quartic factor assume it may be factored as follows:

$x^4+x^2+1=(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+(ab+2) x^2+(a+b)x+1$

Equating coefficients, we find:

$a+b=0$

$ab+2=1$

and so one solution is $(a,b)=(1,-1)$ and we have:

$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$ which means:

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

Method 2:

Let:

$S=x^5+x^4+x^3+x^2+x+1$ and so:

$Sx=x^6+x^5+x^4+x^3+x^2+x=S+x^6-1$ hence:

$S(x-1)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$ thus:

$S=(x+1)(x^2-x+1)(x^2+x+1)$

3. Originally Posted by lookagain
A) Give the completely factored form over the integers of the following polynomial, and

B) demonstrate at least two methods for doing so.

$x^5 + x^4 + x^3 + x^2 + x + 1$

(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)

4. 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.)

5. Here's another way.

Since $x^5+x^4+x^3+x^2+x+1 = \dfrac{x^6-1}{x-1}$ the roots of this polynomial are exactly the set $\{z\in \mathbb{C}-\{1\}\,\,;\,\, z^6=1\}$, i.e. the roots of unity, ignoring the positive real root. They are $e^{\pm i\pi/3},e^{\pm i2\pi/3}, -1$.

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:

$(x-e^{i\pi/3})(x-e^{- i\pi/3}) = x^2-x+1$
$(x-e^{i2\pi/3})(x-e^{- i2\pi/3}) = x^2+x+1$
$x-(-1) = x+1$

There is an abstract algebra/number theoretic variation of the above that can be performed for the general case too.

6. Hello, lookagain

This is a variation of daon's solution.

Demonstrate at least two methods for factoring the following polynomial:

. . . $P(x) \;=\;x^5 + x^4 + x^3 + x^2 + x + 1$

$P(x) \;=\;\dfrac{x^6 - 1}{x-1} \;=\;\dfrac{\overbrace{(x^3)^2 - (1^2)}^{\text{diff. of squares}}}{x-1}$

. . . . .$=\;\dfrac{\overbrace{(x^3-1)}^{\text{diff.of cubes}}\cdot\overbrace{(x^3+1)}^{\text{sum of cubes}}}{x-1}$

. . . . .$=\; \dfrac{(\color{red}{\rlap{/////}}{x-1})(x^2+x+1)(x+1)(x^2-x+1)}{\color{red}{\rlap{/////}}x-1}$

. . . . .$=\; (x+1)(x^2+x+1)(x^2-x+1)$

7. Originally Posted by soroban
Hello, lookagain

This is a variation of daon's solution.

$P(x) \;=\;\dfrac{x^6 - 1}{x-1} \;=\;\dfrac{\overbrace{(x^3)^2 - (1^2)}^{\text{diff. of squares}}}{x-1}$

. . . . .$=\;\dfrac{\overbrace{(x^3-1)}^{\text{diff.of cubes}}\cdot\overbrace{(x^3+1)}^{\text{sum of cubes}}}{x-1}$

. . . . .$=\; \dfrac{(\color{red}{\rlap{/////}}{x-1})(x^2+x+1)(x+1)(x^2-x+1)}{\color{red}{\rlap{/////}}x-1}$

. . . . .$=\; (x+1)(x^2+x+1)(x^2-x+1)$
Originally Posted by MarkFL

Method 2:

Let:

$S=x^5+x^4+x^3+x^2+x+1$ and so:

$Sx=x^6+x^5+x^4+x^3+x^2+x=S+x^6-1$ hence:

$S(x-1)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$ thus:

$S=(x+1)(x^2-x+1)(x^2+x+1)$

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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