# Reordering terms affect the factorization of a square trinomial?

• Mar 22nd 2008, 12:09 PM
norifurippu
Reordering terms affect the factorization of a square trinomial?
I have the square trinomial
$\displaystyle 25 - 10x + x^2$

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

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

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

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

Why do I get different square binomials?
• Mar 22nd 2008, 12:24 PM
Peritus
$\displaystyle \left( {x - 5} \right)^2 = \left[ {\left( { - 1} \right)\left( {5 - x} \right)} \right]^2 = \left( { - 1} \right)^2 \left( {5 - x} \right)^2 = \left( {5 - x} \right)^2$
• Mar 22nd 2008, 12:29 PM
TheEmptySet
They are not so different...
with some factoring we can show they are the same!

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

$\displaystyle (1)[x-5]=(x-5)^2$
• Mar 22nd 2008, 12:47 PM
norifurippu
thanks!
• Mar 22nd 2008, 03:44 PM
mr fantastic
Quote:

Originally Posted by Peritus
$\displaystyle \left( {x - 5} \right)^2 = \left[ {\left( { - 1} \right)\left( {5 - x} \right)} \right]^2 = \left( { - 1} \right)^2 \left( {5 - x} \right)^2 = \left( {5 - x} \right)^2$

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

Even more generally:

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

$\displaystyle (a - b)^n = -(b - a)^n$ if n is odd.
• Mar 22nd 2008, 05:28 PM
norifurippu
Quote:

Originally Posted by mr fantastic
And in general, $\displaystyle (a - b)^2 = (b - a)^2$.

Even more generally:

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

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

Thanks.