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:

$\displaystyle 30x^4-6x^2 = 6x^2(5x^2-1)$

My Solution:

$\displaystyle 30x^4-6x^2 = 3x^2(10x^2-2)$

Both factoring results seem to be correct.

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:

$\displaystyle 30x^4-6x^2 = 6x^2(5x^2-1)$

My Solution:

$\displaystyle 30x^4-6x^2 = 3x^2(10x^2-2)$

Both factoring results seem to be correct.

But $\displaystyle 30x^4-6x^2 = 3x^2(10x^2-2)$ is not factored completely.

Re: Factoring Result Different than Book Example?

In that case, I would say that "$\displaystyle 30x^4- 6x^2= 6x^2(5x^2- 1)$" is also not "factored completely"! The "complete" factoring would be $\displaystyle (2)(3)x^2(5x^2- 1)$.

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

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

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!

Re: Factoring Result Different than Book Example?

Both are correct, but $\displaystyle 6x^2 (5x^2 - 1)$ 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.

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.

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

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

Actually: $\displaystyle 30 \:=\:2\cdot3\cdot5$

. . and **none of you** factored 30 *completely*.

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:

$\displaystyle 30x^4-6x^2 = 6x^2(5x^2-1)$

My Solution:

$\displaystyle 30x^4-6x^2 = 3x^2(10x^2-2)$

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 $\displaystyle \displaystyle \begin{align*} 30x^4 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} 6x^2 \end{align*}$ is $\displaystyle \displaystyle \begin{align*} 6x^2 \end{align*}$, 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...