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So I only half understand this concept, the question is

Simplify $2(x^2+8)^{\frac85} + x \cdot 3(x^2 + 8)^{\frac35} \cdot 9x$

The steps I have so far are

$2(x^2+8)^{\frac85} + 27x^2(x^2 + 8)^{\frac35}$

$(x^2+8)^\frac35[$

After the bracket is where I'm stuck at conceptually. What am I doing with the $27x^2$ and the $2?$ Am I adding them together?

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    can you use $$\LaTeX$$ please? – 2017-01-11
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    Yes, you add them together. – 2017-01-11
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    Sorry I'm not good with the formatting here hold on – 2017-01-11
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    $\frac{2(x^2+8)^8}{5}+\frac{3x(x^2+8)^3}{5}\cdot9x = \frac{(x^2+8)^8}{5}\cdot2+\frac{(x^2+8)^3}{5}\cdot27x^2 = \frac{(x^2+8)^8}{5}\cdot(2+27x^2) = \frac{(x^2+8)^8\cdot(2+27x^2)}{5}$ – 2017-01-11
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    The formatting ruined it, the 3/5 and 8/5 are exponents to the (x^2 + 8) sorry – 2017-01-11
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    @user406105 One editor cleaned up the formats, but changed some of your numbers. I have, (I hope) changed the numbers back to your original post. – 2017-01-11
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    It should be (x^2 + 8) with 3/5 and 8/5 being the exponent for the terms respectively. – 2017-01-11
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    You treat the term "$\frac{(x^2 + 8)}5$" as a unit item, and factor it just as you would a number or a variable. For $2a + 3x*a*9x$ you would do $2a + 3x*a*9x = 2a + 27x^2*a = a[ 2 + 27x^2]$. This is no different: $2\frac{(x^2 + 8)}5 + 3x*\frac{(x^2 + 8)}5*9x = 2\frac{(x^2 + 8)}5 + 27x^2*\frac{(x^2 + 8)}5 = \frac{(x^2 + 8)}5[ 2 + 27x^2]$ – 2017-01-11
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    The formatting is now correct, thanks! – 2017-01-11

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You're not doing anything with $27x^2$ and $2$ just yet. You will keep them inside the brackets, along with some other stuff, as a result of factoring out $(x^2+8)^{\frac{3}{5}}$; you'll simplify the expression in the brackets after that.

Your main goal right now is to factor. Remember that factoring is effectively division (in each term). For example, if you factor $x^2$ out of $x^5$ you will be left with $x^3$, i.e. you would write $x^5=x^2\cdot x^3$. Why? Because $\displaystyle \frac{x^5}{x^2}=x^3$, giving you the remaining factor for $x^5=x^2\cdot(\text{?})$.

Same here. To factor $(x^2+8)^{\frac{3}{5}}$ out of $(x^2+8)^{\frac{8}{5}}$ means to divide: $$\frac{(x^2+8)^{\frac{8}{5}}}{(x^2+8)^{\frac{3}{5}}}=(x^2+8)^{\frac{8}{5}-\frac{3}{5}}=(x^2+8)^1=x^2+8,$$ so your expression simplifies to $$2(x^2+8)^{\frac{8}{5}} + 27x^2(x^2 + 8)^{\frac{3}{5}}=(x^2 + 8)^{\frac{3}{5}}\left[2(x^2+8)+27x^2\right].$$ I hope you can take it from here.

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    So from there, im hoping im correct here the next step would be (x2+8)^(3/5)[2x^2+ 16 +27x^2] -> (x2+8)^(3/5)[29x^2 + 16] and from there im done correct? – 2017-01-11
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    @user406105: Yes, that's right! – 2017-01-11
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    If you don't mind me asking for help on a second problem $28x^{\frac12}(x^2 + x) - 12x^{\frac32} - 20x^{\frac12}$ Now this has 3 terms if im not mistaken and we only did practice problems with 2 terms. How do I simplify these 3 terms easily, and how do i treat the fractional exponents being right next to the x and not with a parenthesis. – 2017-01-11
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    Same thing: factor $x^{\frac{1}{2}}$ out of all three terms to get $x^{\frac{1}{2}}[\ldots-\ldots-\ldots]$. And use the same technique for each factoring. – 2017-01-11
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    $x^{\frac12}[28(x2+x)−12.. - 20]$ I believe this is correct, with the only thing me missing is what to do with the 12 as it has a 3/2 fraction, how does that work with the distribution? – 2017-01-11
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    Same thing: to factor out means to divide, so $x^{\frac{3}{2}}/x^{\frac{1}{2}}=x^{\frac{3}{2}-\frac{1}{2}}=x^1=x$. – 2017-01-11