Factoring 6x(x^2+1)^2(x^3-1) from an algebraic expression

[courtrigrad]

Hello


I do not understand how to get from:

$$(x^2+1)^3 6x^2(x^3 - 1) + (x^3-1)^2 6x(x^2+1)^2$$

to $$6x(x^2+1)^2(x^3-1)(2x^3 + x - 1)$$

I tried factoring the $$6x$$ but have had no luck.

Any help is appreciated!

Thanks

 

[MathStudent]

try factoring 6x rather than 6x^2

 

[Hurkyl]

Well, you can't factor 6x^2, because the second term doesn't have an x^2!

Here, you're just trying to apply the distributive rule in reverse:

ab + ac = a(b + c)

What you need to do is identify the factors that appear in both terms. (e.g. x appears in both terms, but not x^2), and that's what you factor out. (that's the a).

 

[dextercioby]

You can't factor $6x^{2}$,without getting something really ugly...

Pay attention with the calculations.The fact that u know the result already might help u if u don't see means of factorization...

Daniel.

 

[Curious3141]

Hello


I do not understand how to get from:

$$(x^2+1)^3 6x^2(x^3 - 1) + (x^3-1)^2 6x(x^2+1)^2$$

to $$6x(x^2+1)^2(x^3-1)(2x^3 + x - 1)$$

I tried factoring the $$6x^2$$ but have had no luck.

Any help is appreciated!

Thanks

$$(x^2+1)^3 6x^2(x^3 - 1) + (x^3-1)^2 6x(x^2+1)^2 = (6x)(x^3 - 1)(x^2 + 1)^2\left( (x(x^2 + 1) + (x^3 - 1) \right) = 6x(x^2+1)^2(x^3-1)(2x^3 + x - 1)$$

Just group the terms and play with the algebra.

 

[Hurkyl]

Pfft, don't do the work for him -- one learns more when they do the work, rather than observing someone else's work.

 

[Curious3141]

Pfft, don't do the work for him -- one learns more when they do the work, rather than observing someone else's work.

Point taken

 

[dextercioby]

So much for my advice...

Not to mention u messed up the page layout...

Daniel.

 

[courtrigrad]

(he does the work (singular) )

anyway i worked it out (like 1:20 AM here)

Thanks all

 

[Curious3141]

Not to mention u messed up the page layout...

Daniel.

Then set your browser differently.

 

[futb0l]

Up the resolution, it's just fine on 1280x1024 :P