How to show $x^4 - 1296 = (x^3-6x^2+36x-216)(x+6)$

Question

How to get this result: $x^4-1296 = (x^3-6x^2+36x-216)(x+6)$?

It is part of a question about finding limits at mooculus.

Answer 1

Hints: $1296=(-6)^4$ and $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1})$.

Answer 2

Here's a harder and similar way, but it still makes sense if you don't have what Julien said memorized.

$$x^4 - 1296 = x^4 - 6^4 \Rightarrow x^4 = 6^4$$Now two roots that seem obvious are $6$ and $-6$. But $x^4 = x^4 \cdot 1 = x^4 \cdot i^4 = (xi)^4$. So two more roots are $6i$ and $-6i$.

Hence, the factorization is $$\begin{aligned}&(x - 6)(x + 6)(x+6i)(x - 6i) \\ =& (x-6)(x+6)(x^2+36) \\ =& (x+6)(x^3 - 6x^2 +36x-216)\end{aligned}$$

Answer 3

$$\begin{array}{r|rrrr|r} & 1 & 0 & 0 & 0 & -1296 \\ -6 & & -6 & 36 & -216 & 1296 \\ \hline & 1 & -6 & 36 & -216 & 0 \end{array} $$

This shows that $$ x^4-1296=(x+6)(x^3-6x^2+36x-216) $$

Of course you can also use

$$ a^4 - b^4 = (a-b)(a^3 + a^2b + ab^2 + b^3) $$ with $a=x$ and $b=-6$.

Answer 4

A bit more roundabout way:

$x^4 - 1296 \ $ is a difference of two squares, as is $ \ x^2 - 36 \ $, so

$$(x^4 - 1296) \ = \ [ \ (x^2)^2 \ - \ (36^2) \ ] \ = \ (x^2 + 36) \ \cdot (x^2 - 36 ) $$

$$= \ (x^2 + 36) \cdot (x - 6) \cdot (x + 6) \ = \ (x^3 \ + \ 36x \ - \ 6x^2 - 216 ) \cdot (x + 6) \ . $$

EDIT: I missed that Πάρτη Κοηλί had done this going down to complex roots; this version "keeps things real"...

All of the answers I see at the present are fine -- there isn't any getting around that one needs to recognize that 1296 is $ \ 6^4 \ $ , or otherwise needs to find enough factors of 1296 to see a way to "break down" the polynomial.

Answer 5

It is a basic result of field theory that if a binomial $\, x^n - c\,$ is reducible then it has a proper factor given by the Factor Theorem or difference of squares, of one of the following two forms

$$\begin{eqnarray} \text{Factor Theorem}\!:\ \ & x^n - a^n &=\,& (x\ -\ a)\,(\cdots) \\ \Rightarrow\ \ {\rm Form_1}\!:\ \ &(x^k)^p\!-a^p &=\,& (x^k- a)\, (\cdots)\quad \text{for prime}\ \ p\mid n \\ \Rightarrow\ \ {\rm Form_2}\!:\ \ & x^{4k}\! + 4a^4 &=\,& (x^{2k}\!+2a^2)^2 - (2ax^k)^2 = \ \cdots \end{eqnarray}$$

So $\, x^4\! - 1296\,$ reducible $\,\Rightarrow\, 1296 = a^2,\,a^4,\,{\rm or}\,\ {-}4a^4.\,$ $\,1296 = 4\cdot 324 = 4\cdot 4\cdot 81 = 2^4\cdot 3^4,\,$ hence we obtain factors of Form$_1,\,$ viz. $\, x^4-(\pm a)^4 = (x\mp a)\,(\cdots)$

A proof of said theorem can be found in Karpilovsky, Topics in Field Theory, Theorem 8.1.6.