# Having a difficult time with more advanced distributing in algebra?

• November 16th 2011, 04:12 PM
Mikusboi
Having a difficult time with more advanced distributing in algebra?
Hello, I'm in Business with Algebra and Calculus and I know the Calculus pretty well but it's this particular type of distributing that I'm having trouble with. For instance..
(x^4-2) * 3(x^2+2)^2 *2x + (x^2+2)^3 * 4x^3

There is just so many numbers I can't figure out which numbers and values I should take out

2x(x^2+2)^2[3x^4 - 6 + 2x^4 + 4x^2]

Then:
2x(x^2 +2)^2 * (5x^4 + 4x^2 - 6)

I just need to figure out how to do this then I think I will be good for my exam.
• November 16th 2011, 04:41 PM
Quacky
Re: Having a difficult time with more advanced distributing in algebra?
$\underbrace{(x^4-2)\times{3(x^2+2)^2}\times{2x}} + (x^2+2)^3\times{4x^3}$
In the context I will be using, this is going to be seen as one term, as everything is connected through multiplication. Everything after the $+$ that follows constitutes the second term.

Look for common factors to both terms of the expression. I've highlighted some below:

$(x^4-2)\times{3(x^2+2)^2}\times{{\color{red}2x}} + (x^2+2)^3\times{{\color{red}4x^3}}$

Both terms in red are multiples of $2x$ - the latter is $2\times{2}\times{x}\times{x}\times{x}$

So we can take this out as a factor:

$=2x[$

And then we just need to divide each term by $2x$ once - almost everything stays the same, and we have:

$=2x[(x^4-2)\times{3(x^2+2)^2} + (x^2+2)^3\times{2x^2}]$ as the only terms that have changed are the terms in red.

We can then look for another common factor - and indeed there is one.

$=2x[(x^4-2)\times{3{\color{red}(x^2+2)^2}} + {\color{red}(x^2+2)^3}\times{2x^2}]$

We can take out a factor of $(x^2+2)^2$ as this is the highest power of $(x^2+2)$ that is common to both terms.

$=2x(x^2+2)^2[$

And again, we divide through each term by the common factor once.

$=2x(x^2+2)^2~[(x^4-2)\times{3} + (x^2+2)\times{2x^2}]$

Notice that the last term still has $(x^2+2)^1$ remaining, because it was initially to the power 3 and only the second power was common to both terms.

Now it's a case of simplifying what's left in the square brackets - hopefully you'll be okay with that.
• November 16th 2011, 05:18 PM
Mikusboi
Re: Having a difficult time with more advanced distributing in algebra?
Thanks for the very informative answer, I think I'm beginning to understand it now. The plus sign separating the two terms explained a lot. I think I just have to practice these sort of problems some more now :). BTW, how did you show your answer in handwriting-like form?
• November 16th 2011, 05:25 PM
Quacky
Re: Having a difficult time with more advanced distributing in algebra?
I hope it helped. To type the mathematical notation requires an understanding of latex, but basically, using the [Tex][\tex] tags at the top right of the page, you can do the following:

[TEX]x^2+7x+9[\TEX] gives $x^2+7x+9$

The codes for other symbols vary, but you'll get used to them in time.
• November 16th 2011, 05:49 PM
Mikusboi
Re: Having a difficult time with more advanced distributing in algebra?
Ah I see, very cool. I'll use this in the future.