# Factoring Result Different than Book Example?

• Jun 25th 2012, 11:26 AM
allyourbass2212
Factoring Result Different than Book Example?
It seems there are times where its possible to arrive at different factoring results. For instance view this problem from the book:

Books Solution:
$30x^4-6x^2 = 6x^2(5x^2-1)$

My Solution:
$30x^4-6x^2 = 3x^2(10x^2-2)$

Both factoring results seem to be correct.
• Jun 25th 2012, 12:07 PM
Plato
Re: Factoring Result Different than Book Example?
Quote:

Originally Posted by allyourbass2212
It seems there are times where its possible to arrive at different factoring results. For instance view this problem from the book:

Books Solution:
$30x^4-6x^2 = 6x^2(5x^2-1)$

My Solution:
$30x^4-6x^2 = 3x^2(10x^2-2)$

Both factoring results seem to be correct.

But $30x^4-6x^2 = 3x^2(10x^2-2)$ is not factored completely.
• Jun 25th 2012, 12:23 PM
HallsofIvy
Re: Factoring Result Different than Book Example?
In that case, I would say that " $30x^4- 6x^2= 6x^2(5x^2- 1)$" is also not "factored completely"! The "complete" factoring would be $(2)(3)x^2(5x^2- 1)$.

Of course that's is specifically "factoring with integer coefficients". Without that restriction, we could also factor as $(2)(3)x^2(x\sqrt{5}- 1)(x\sqrt{5}+ 1)$.

Or as $(\sqrt{2})(\sqrt{2})(\sqrt{3})(\sqrt{3})x^2(x\sqrt {5}- 1)(x\sqrt{5}+ 1)$

There are, in fact, an infinite number of ways to factor that. Unfortunately, this text box is not large enough for me to show all of them!
• Jun 25th 2012, 01:06 PM
richard1234
Re: Factoring Result Different than Book Example?
Both are correct, but $6x^2 (5x^2 - 1)$ is more completely factored.

There are other examples of polynomials that have several different factorizations, all correct, but don't easily follow from another factorization.
• Jun 25th 2012, 01:27 PM
Soroban
Re: Factoring Result Different than Book Example?
Hello, allyourbass2212!

Quote:

It seems there are times where its possible to arrive at different factoring results.
. . This is not true.

By definition, "factor" always means "factor completely".

Otherwise, we might have this disagreement.

Problem: factor 30.

$\begin{array}{ccccccc}\text{You could claim that:} & 30 &=& 2\cdot 15 \\ \text{Someone else could claim:} & 30 &=& 3\cdot10 \\ \text{Yet another could claim:} & 30 &=& 5\cdot6 \end{array}$

And you could argue (forever) over who is right.

Actually: $30 \:=\:2\cdot3\cdot5$
. . and none of you factored 30 completely.
• Jun 25th 2012, 10:09 PM
Prove It
Re: Factoring Result Different than Book Example?
Quote:

Originally Posted by allyourbass2212
It seems there are times where its possible to arrive at different factoring results. For instance view this problem from the book:

Books Solution:
$30x^4-6x^2 = 6x^2(5x^2-1)$

My Solution:
$30x^4-6x^2 = 3x^2(10x^2-2)$

Both factoring results seem to be correct.

To avoid confusing the OP, we should point out that in the context of THIS question, the book appears to be asking to factorise this particular expression COMPLETELY OVER THE RATIONALS.

The highest rational common factor of \displaystyle \begin{align*} 30x^4 \end{align*} and \displaystyle \begin{align*} 6x^2 \end{align*} is \displaystyle \begin{align*} 6x^2 \end{align*}, that is why it has been taken out.

Notice that in your solution, there is still a rational common factor of 2 inside your brackets...