I apologize if this question already exists, but it was quite difficult to word.
In my math class today, we learned how to factor a difference of two perfect cubes. One of our practice questions was:
factor $x^6 - 64$
almost everyone other than me (including my teacher, as he was rushing) ended up with:
$(x^2 - 4)(x^4 + 4x^2 + 16)$
I pointed out two things:
1:
$(x^6 - 64)$ was also a difference of two perfect squares, and we could factor it into this:
$(x^3 + 8)(x^3 - 8)$
Which could then be factored further into:
$(x + 2)(x^2 - 2x + 4)(x - 2)(x^2 + 2x + 4)$
and 2:
the $(x^2 - 4)$ was also a difference of two perfect squares, and we could factor further:
$(x + 2)(x - 2)(x^4 + 4x^2 + 16)$
Now for my question:
How are $(x + 2)(x^2 - 2x + 4)(x - 2)(x^2 + 2x + 4)$ and $(x + 2)(x - 2)(x^4 + 4x^2 + 16)$ equivalent?
The teacher didn't seem to know off the top of his head, and I can't figure it out after trying for a half an hour. I've heard that you can factor the sum of two perfect squares with imaginary numbers, so maybe i can do something there to help explain this?
edit: fixed instances of "- 16" to "+ 16", and "4x" to "4x^2" (thanks for pointing that out)
$\endgroup$ 34 Answers
$\begingroup$It's a very common mistake (*) to think that, since $t^2+4t+16$ is irreducible over the reals, also the polynomial obtained with $t=x^2$ is irreducible.
Over the reals, every polynomial of degree $>2$ is reducible, so also $x^4+4x^2+16$ is reducible.
The trick here is to push in a difference of squares: if you add and subtract $4x^2$, you get $$ x^4+8x^2+16-4x^2=(x^2+4)^2-(2x)^2=(x^2+4-2x)(x^2+4+2x) $$ so you get the same factorization as with the other method.
Another funny example is $x^4+1$: $$ x^4+1=x^4+2x^2+1-2x^2= (x^2+1)^2-(\sqrt{2}\,x)^2= (x^2+1-\sqrt{2}\,x)(x^2+1+\sqrt{2}\,x) $$ which is the full factorization over the reals, since those degree $2$ polynomials have no real roots.
(*) Even among teachers, unfortunately.
$\endgroup$ $\begingroup$There are two errors in the factorization
$$x^6-64=(x^2-4)(x^4+4x-16)$$
The constant term in the quartic should be $+16$ instead of $-16$, and the linear term, $4x$, should be a quadratic term, $4x^2$. That is, the correct factorization is
$$x^6-64=(x^2-4)(x^4+4x^2+16)$$
This now agrees with the factorization $(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)$ since $(x-2)(x+2)=x^2-4$ and
$$\begin{align} (x^2-2x+4)(x^2+2x+4)&=(x^2+4-2x)(x^2+4+2x)\\ &=(x^2+4)^2-(2x)^2\\ &=x^4+8x^2+16-4x^2\\ &=x^4+4x^2+16 \end{align}$$
$\endgroup$ $\begingroup$A complete factorization of $x^6 - y^6$ can be performed as follows: $$\begin{align*} x^6 - y^6 &= (x^3)^2 - (y^3)^2 \\ &= (x^3 - y^3)(x^3 + y^3) \\ &= (x-y)(x^2 + xy + y^2)(x+y)(x^2 - xy + y^2). \end{align*}$$ The quadratic factors are irreducible in the ring of polynomials with rational coefficients $\mathbb Q[x]$ (more informally, they don't factor "nicely"). We can recombine these factors in carefully chosen ways to get expressions that might have resulted from an incomplete factorization; e.g., if we let $a = x^2 + y^2$ and $b = xy$, then $$\begin{align*} (x^2 + xy + y^2)(x^2 - xy + y^2) &= (a+b)(a-b) \\ &= a^2 - b^2 \\ &= (x^2+y^2)^2 - (xy)^2 \\ &= x^4 + 2x^2 y^2 + y^4 - x^2 y^2 \\ &= x^4 + x^2 y^2 + y^4. \end{align*}$$ And if we combine $(x-y)(x+y)$ we get $x^2 - y^2$; thus we recover the incomplete factorization of $x^6 - y^6$ as a difference of cubes $$(x^2)^3 - (y^2)^3 = (x^2 - y^2)(x^4 + x^2 y^2 + y^4).$$ There are of course other ways to combine these factors, each leading to a different expression, but all are algebraically equivalent.
$\endgroup$ 1 $\begingroup$"How are $(x+2)(x^2−2x+4)(x−2)(x^2+2x+4)$ and $(x+2)(x−2)(x^4+4x^2+16)$ equivalent?"
Because $(x^2−2x+4)(x^2+2x+4)=x^4+4x^2+16$
===
Nother way of looking at it.
$a^3-b^3=(a-b)(a^2+ab+b^2) $
$a^2-b^2=(a-b)(a+b) $
So therefore $a^6-b^6=(a^2-b^2)(a^4+a^2b^2+b^4) =(a-b)(a+b) (a^4+a^2b^2+b^4)$
Yet, $a^6-b^6=(a^3-b^3)(a^3+b^3)=(a-b)(a^2+ab+b^2)(a^3+b^3) $
So it must be that $ (a+b) (a^4+a^2b^2+b^4)= (a^2+ab+b^2)(a^3+b^3) $ somehow.
So $(a+b)|(a^2+ab+b^2) $ or $(a+b)|(a^3+b^3) $.
$a^2+ab+b^2=a (a+b)+b^2$ which doesn't seem to work.
$a^3+b^3=a^2 (a+b)-a^2b+b^3=a^2(a+b)-ab (a+b)+ab^2+b^3=a^2 (a+b)-ab (a+b)+b^2 (a+b)-b^3+b^3=(a+b)(a^2-ab+b^2) $
So now we must have $ (a+b) (a^4+a^2b^2+b^4)= (a+b)(a^2+ab+b^2)(a^2-ab+b^2)$ and so it must be that:
$a^4+a^2b^2+b^4=(a^2+ab+b^2)(a^2-ab+b^2)$
And if we try to multiply it out, we see that it is so.
I don't blame the students for not seeing this, but the teacher ought to have recognize that it had to happen.
$\endgroup$ 2
