Factorise: $a^5+a^4+a^3+a^2+a+ 1$

Question

Honestly I have no idea where to start with this one

If i take out the 'a' first, where do i go after that

Can you think of any other ways to go about this question?

Answer 1

HINT

To start with, I would recommend to proceed as follows

\begin{align*} a^{5} + a^{4} + a^{3} + a^{2} + a + 1 & = a^{3}(a^{2} + a + 1) + (a^{2} + a + 1)\\\\ & = (a^{3} + 1)(a^{2} + a + 1) \end{align*}

Then apply the sum of cubes formula. Can you take it from here?

Answer 2

I'd start with $$(a-1)\left(a^5+a^4+a^3+a^2+a+ 1\right)=a^6-1$$

And then you have $$a^6-1=\left(a^3+1\right)\left(a^3-1\right)$$

You can keep factoring those two factors. And then divide away the $(a-1)$ from where we started.

Answer 3

We have
$a^5+a^4+a^3+a^2+a+ 1=\dfrac{a^6-1}{a-1}=\dfrac{(a^3-1)(a^3+1)}{a-1}=\dfrac{(a-1)(a^2+a+1)(a+1)(a^2-a+1)}{a-1}=(a + 1) (a^2 - a + 1) (a^2 + a + 1)$

Answer 4

The expression $=\frac{a^6-1}{a-1}$. Now $a^6-1=0$ has six roots $r_n=e^{\frac{n\pi i}{3}}$, for $0\le n\le 5$. The factors you want are $a-r_n$, for $1\le n\le 5$. Note that $r_0=1$.

Answer 5

You cannot factorise this expression completely (i.e. linear factors) in terms of real numbers.

You can do one of the following:

Write $a^5 + a^4 + a^3 + a^2 + a + 1 = \frac{a^6-1}{a-1}$. This is not factorisation, but it may be considered a simplification, and may be useful to you.

Or: $a^5 + a^4 + a^3 + a^2 + a + 1 = a^3(a^2+a+1) + a^2+a+1 = (a^3+1)(a^2+a+1) = (a+1)(a^2-a+1)(a^2+a+1)$, which is the simplest real number factorisation you can achieve, as neither of the two degree two (quadratic) factors has a real root.

Or, finally, the full linear factorisation in terms of complex numbers: $a^5 + a^4 + a^3 + a^2 + a + 1 = (a-z_1)(a-z_2)(a-z_3)(a-z_4)(a-z_5)$, where each $z_k$ represents a distinct complex sixth root of unity (one). $z_k = e^{i\frac{k\pi}{3}} = \cos \frac{k\pi}{3} + i\sin \frac{k\pi}{3}, k = 1, 2, 3, 4, 5$.

Answer 6

Grouping the terms by two and then by three, one obtains $$ a^5 + a^4 + a^3 + a^2 + a + 1 = (a^5 + a^4) + (a^3 + a^2) + (a + 1) = (a^5 + a^4 + a^3) + (a^2 + a + 1) $$ Since $a^5 + a^4 = a^4(a+1)$, $a^3 + a^2 = a^2(a+1)$ and $a^5 + a^4 + a^3 = a^3(a^2 + a + 1)$, one gets $$ a^5 + a^4 + a^3 + a^2 + a + 1 = (a^4 + a^2 + 1)(a+1) = (a^3+1)(a^2+a+1) $$ It follows that $a^2 + a + 1$ divides $a^4 + a^2 + 1$ and $a+1$ divides $a^3 + 1$ (you could also see this directly by noting that $-1$ is a root of $a^3 +1$). Completing the divisions yields $a^3 + 1 = (a+1)(a^2 -a + 1)$ and $a^4 + a^2 + 1 = (a^2 + a + 1)(a^2 -a + 1)$.