MHB Factoring y^4+y^2+1: Other Ways to Approach

[Dethrone]

How do you factor $$y^4+y^2+1$$?
I know that if you regroup them like this:
$$(y^4+2y^2+1)-y^2$$, factoring becomes apparent:
$$(y^2 + 1 + y) (y^2 + 1 - y)$$

Is there any other way to factor this? It's just a pain to have to try out these "tricks" every time I try to factor something.

 

[Deveno]

How do you factor $$y^4+y^2+1$$?
I know that if you regroup them like this:
$$y^4+2y^2+1-y^2$$, factoring becomes apparent:
$$(y^2 + 1 + y) (y^2 + 1 - y)$$

Is there any other way to factor this? It's just a pain to have to try out these "tricks" every time I try to factor something.

Unfortunately, no, factoring is what as known as a "hard problem", because it is often no easy task to determine if a factorization even exists. So a "bag of tricks" is one's best friend.

Here, we catch a lucky break, the coefficients of $y^4 + y^2 + 1$ are all integers, which means that if it DOES factor, the coefficients of the factors will ALSO be integers. It is not hard to see we have no linear factors, since no integer choice for $y$ will make the polynomial 0.

So if we have any factors at all, they are quadratic factors, and we have two choices:

$y^4 + y^2 + 1 = (y^2 + ay - 1)(y^2 + by - 1)$, or:

$y^4 + y^2 + 1 = (y^2 + ay + 1)(y^2 + by + 1)$

Let's look at the first one. We have:

$(y^2 + ay - 1)(y^2 + by - 1) = y^4 + (a+b)y^3 + (ab - 2)y^2 - (a+b)y + 1$

Comparing this to $y^4 + y^2 + 1$, we see we must have:

$a+b = 0$
$ab = 3$

Thus $ab = -a^2 = 3$, which is impossible for an integer.

So we turn to the second one:

$(y^2 + ay + 1)(y^2 + by + 1) = y^4 + (a+b)y^3 + (ab + 2)y^2 + (a+b)y + 1$, which tells us:

$a+b = 0$
$ab = -1$

This, in turn, leads to $ab = -a^2 = -1$, so $a^2 = 1$, which has the solutions $a = \pm 1$. If we take $a = 1$, we get $b = -1$, and vice-versa, so both choices lead to the same factorization:

$y^4 + y^2 + 1 = (y^2 + y + 1)(y^2 - y + 1)$.

Sometimes, an analysis like this is all one can do, and the resulting system of equations can be challenging to solve (they are typically non-linear). This becomes increasingly true of polynomials of higher degree (like more than 6), which are quite difficult to factor.

Of course your "trick" is quicker than what I did. That's why such "short-cuts" continue to be useful.

 

[Dethrone]

Thanks so much for the insightful response, I have learned much from it!
Will there be a case where even quadratic factors don't exist, and we have to use higher order/degree factors?

 

[Deveno]

Yes. For example, it is not possible to factor:

$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ using only rational numbers.

 

[Dethrone]

Thanks again!

$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ using only rational numbers.

$$=-((-1)^{2/7}-x) ((-1)^{4/7}-x) (-x+(-1)^{5/7}-(-1)^{4/7}+(-1)^{3/7}-(-1)^{2/7}+(-1)^{1/7}-1) (x+(-1)^{1/7}) (x+(-1)^{3/7}) (x+(-1)^{5/7})$$ (Nod)

P.S: How can you factor that?

Spoiler

Not factoring, but I realize that the above expression can be written as: $$\frac{x^7-1}{x-1}$$

 

[Deveno]

Thanks again!

$$=-((-1)^{2/7}-x) ((-1)^{4/7}-x) (-x+(-1)^{5/7}-(-1)^{4/7}+(-1)^{3/7}-(-1)^{2/7}+(-1)^{1/7}-1) (x+(-1)^{1/7}) (x+(-1)^{3/7}) (x+(-1)^{5/7})$$ (Nod)

P.S: How can you factor that?

Spoiler

Not factoring, but I realize that the above expression can be written as: $$\frac{x^7-1}{x-1}$$

Yes, you can factor ANY polynomial if you "enlarge the field" enough.

But in your example, $1^{1/7}$ isn't "1", but a primitive 7-th root of unit:

$\cos\left(\dfrac{2\pi}{7}\right) + i\sin\left(\dfrac{2\pi}{7}\right)$

which isn't even a real number, but a complex one.

 

[kaliprasad]

this is a very special case of $y^{2n} + y^n+ 1 $ where n is not multiple of 3 . In this case it is one 1. so w and $w^2$ are zeroes of the polynomial hence $y^2+y+1$ is a factor and by dividing we get the 2nd factor(note w is cube root of 1)

 

[Dethrone]

How did you conclude that $w$ and $w^2$ are the roots?

 

[Deveno]

Suppose that $\omega$ is a complex number such that:

$\omega^3 = 1$ and $\omega \neq 1$.

It follows that $(\omega^2)^3 = (\omega^3)^2 = 1^2 = 1$ as well.

So both $\omega,\omega^2$ are roots of $x^3 - 1$. Since 1 is a root as well, we have:

$x^3 - 1 = (x - 1)(x - \omega)(x -\omega^2)$, since a cubic can only have 3 roots at most.

Hence:

$(x - \omega)(x - \omega^2) = \dfrac{x^3 - 1}{x - 1} = x^2 + x + 1$.

Is there actually such a complex number? Yes, we may take:

$\omega = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}$

$=\cos\left(\dfrac{2\pi}{3}\right) + i\sin\left(\dfrac{2\pi}{3}\right)$

$= e^{2\pi i/3}$ (1/3-rd of the way around the unit circle in the complex plane).

kaliprasad's trick is another one you can "put in your bag".

 

[Dethrone]

Thanks!

Spoiler

My bag is still quite empty, anymore useful factoring tricks? :D I know the one for $x^2+y^2$, where $2xy$ is a perfect square.

 

[I like Serena]

Here, we catch a lucky break, the coefficients of $y^4 + y^2 + 1$ are all integers, which means that if it DOES factor, the coefficients of the factors will ALSO be integers.

Erm... how would we factor $x^2-x-1$?

 

[soroban]

Hello, Rido12!

Here is another method.
It too requires a "trick".

Factor $$x^4+x^2+1$$

Multiply by (x^2-1)

\begin{array}{ccc} (x^2-1)(x^4+x^2+1) &amp;=&amp; x^6 - 1\qquad\qquad\qquad\qquad\qquad\qquad\quad \\<br /> (x^2-1)(x^4+x^2+1) &amp;=&amp; (x^3-1)(x^3+1) \qquad\qquad\qquad\qquad\quad \\<br /> (x^2-1)(x^4+x^2+1) &amp;=&amp; (x-1)(x^2+x+1)(x+1)(x^2-x+1) \\<br /> (x^2-1)(x^4+x^2+1) &amp;=&amp; (x-1)(x+1)(x^2+x+1)(x^2-x+1) \\<br /> (x^2-1)(x^4+x^2+1) &amp;=&amp; (x^2-1)(x^2+x+1)(x^2-x+1) \qquad \end{array}Divide by (x^2-1)\!:

. . x^4 + x^2 + 1 \;=\;(x^2+x+1)(x^2-x+1)

 

[Dethrone]

I also noticed something cool that I'd like to share, let's all add something interesting to this thread!

$$x^n -1 = (x-1)(\sum_{i=1}^{n}x^{n-i})$$

I'm really not sure if I notated it correctly, but:
$$x^2-1=(x-1)(x+1)$$
$$x^3-1=(x-1)(x^2+x+1)$$
$$x^4-1=(x-1)(x^3+x^2+x+1)$$
$$x^5-1=(x-1)(x^4+x^3+x^2+x+1)$$

Also, soroban, I do see how your trick works, but can it be generalized? Why $x^2-1$? Is it because it multiplies nicely to $x^6-1$?