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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!

Asaf Karagila
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aman singh
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4 Answers4

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$(x-1)$ 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 $-1$).

JMoravitz
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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.

dxiv
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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.

David
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  • "so is any non-zero constant times $x-1$" in the case of $\Bbb R[x]$, but not in terms of $\Bbb Z[x]$. $(2x-2)$ does not divide evenly into $x^2-1$ in $\Bbb Z[x]$ for example. – JMoravitz Apr 30 '18 at 04:48
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$$x^2-1 =(x-1)(x+1) = (1-x)(-1-x)$$

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.

Mohammad Riazi-Kermani
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