Reordering terms affect the factorization of a square trinomial?

**norifurippu** · March 22nd 2008, 12:09 PM

I have the square trinomial
$25 - 10x + x^2$

If I factor it I get:
$(5 - x)^2$

but If I rearrange the addends in an order in which literal addend are to precede numerial addends,
$x^2 -10x + 25$ ,

and factor it I get:
$(x-5)^2$

aren't $25 - 10x + x^2$ and $x^2 -10x + 25$ equal in value?

Why do I get different square binomials?

**Peritus** · March 22nd 2008, 12:24 PM

$<br /> \left( {x - 5} \right)^2 = \left[ {\left( { - 1} \right)\left( {5 - x} \right)} \right]^2 = \left( { - 1} \right)^2 \left( {5 - x} \right)^2 = \left( {5 - x} \right)^2$

**TheEmptySet** · March 22nd 2008, 12:29 PM

with some factoring we can show they are the same!

$(5-x)^2=[(-1)(-1)5+(-1)x]^2=[(-1)((-1)5+x)]^2=(-1)^2[(-1)5+x]^2=$

$(1)[x-5]=(x-5)^2$

**norifurippu** · March 22nd 2008, 12:47 PM

thanks!

**mr fantastic** · March 22nd 2008, 03:44 PM

Originally Posted by Peritus

$<br /> \left( {x - 5} \right)^2 = \left[ {\left( { - 1} \right)\left( {5 - x} \right)} \right]^2 = \left( { - 1} \right)^2 \left( {5 - x} \right)^2 = \left( {5 - x} \right)^2$

And in general, $(a - b)^2 = (b - a)^2$ .

Even more generally:

$(a - b)^n = (b - a)^n$ if n is even

$(a - b)^n = -(b - a)^n$ if n is odd.

**norifurippu** · March 22nd 2008, 05:28 PM

Originally Posted by mr fantastic

And in general, $(a - b)^2 = (b - a)^2$ .

Even more generally:

$(a - b)^n = (b - a)^n$ if n is even

$(a - b)^n = -(b - a)^n$ if n is odd.

Thanks.

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Reordering terms affect the factorization of a square trinomial?

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