Why is HCF of $x^2-1$ and $x-1$ is $x-1$, "why can't it be $1-x$?"

Question

I've been to different site and tried to find it using Mathematica also, but everywhere they put the answer $x-1$ before me. I even tried to find for $1-x^2$ and $1-x$, then also I got the answer $x-1$.

Can anyone explain please!

Answer 1

$(x-1)$ and $(1-x)$ are considered "the same" in terms of factors in $\Bbb Z[x]$ as well as in $\Bbb R[x]$ since they differ only by multiplication by a unit (in this case $-1$).

Answer 2

A $\gcd$ (or $\operatorname{hcf}$) is only defined up to associates in the respective ring. For real polynomials, the usual convention is to define the $\gcd$ to be the monic polynomial among those, since the units are the (non-zero) constant polynomials.

Answer 3

You can define a HCF of polynomials $f$ and $g$ to be a polynomial $h$ with the two properties

• $h\mid f$ and $h\mid g$;
• if $p\mid f$ and $p\mid g$ then $p\mid h$.

In this case $x-1$ and $1-x$ are both correct, and in fact so is any non-zero constant times $x-1$.

However some people add a third property to the definition,

• $h$ is monic, that is, the leading coefficient is $1$.

In this case $x-1$ is the only correct answer.

Answer 4

$$x^2-1 =(x-1)(x+1) = (1-x)(-1-x)$$

Thus 1-x and x-1 are both factors.

In $\Bbb R[x]$, you may as well multiply these factors by any real number to get factors.

For example $$ x^2-1 =(1/2)(2x-2)(x+1)$$ so $2x-2$ is also a factor.