# challenge question -- Factor the polynomial completely

#### lookagain

##### Senior Member
Edit:

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

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

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#### 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)(x[SUP]4[/SUP]+x[SUP]2[/SUP]+1) → (x+1)(x[SUP]4[/SUP]+2x[SUP]2[/SUP]+1 - x[SUP]2[/SUP]) → (x+1)(x[SUP]2[/SUP]+1+x)(x[SUP]2[/SUP]+1-x)

or

(x[SUP]3[/SUP]+1)(x[SUP]2[/SUP]+x+1) →(x+1)(x[SUP]2[/SUP]-x+1)(x[SUP]2[/SUP]+x+1)

#### lookagain

##### Senior 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.)

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#### daon2

##### Full 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

##### Senior 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)