Factoring Result Different than Book Example?

**allyourbass2212** · June 25th 2012, 10:26 AM

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.

**Plato** · June 25th 2012, 11:07 AM

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.

**HallsofIvy** · June 25th 2012, 11:23 AM

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!

**richard1234** · June 25th 2012, 12:06 PM

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.

**Soroban** · June 25th 2012, 12:27 PM

Hello, allyourbass2212!

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.

**Prove It** · June 25th 2012, 09:09 PM

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...

Thread: Factoring Result Different than Book Example?

LinkBack

Thread Tools

Display

Factoring Result Different than Book Example?

Re: Factoring Result Different than Book Example?

Re: Factoring Result Different than Book Example?

Re: Factoring Result Different than Book Example?

Re: Factoring Result Different than Book Example?

Re: Factoring Result Different than Book Example?

Similar Math Help Forum Discussions

Need to Prove Almost Everywhere Result

how can u get this result?

A result of differentiability

prove this result please

factoring (common factoring complex trinomials)

Search Tags