HELP MATH?

4 x^(-3/5) + 8 x^(1/5)
= [4 / x^(3/5)] + 8 x^(1/5)
= [4 / x^(3/5)] + 8 x^(2/5) / x^(3/5)
= [4 + 8 x^(2/5)] / x^(3/5)
= 4 [1 + 2 x^(2/5)] / x^(3/5)

4x^(-3/5) + 8x^(1/5)

Before we do that, let's review how we factor normal polynomials to remind ourselves of the steps needed to handle this:

x² + x

x is the common factor, so we can pull it out.

Then every term inside gets divided by x to get the resulting factor:

x(x + 1)

Division of two numbers of the same base is the same as subtracting exponents.

Now let's go back to your problem:

4x^(-3/5) + 8x^(1/5)

4 is a common factor, so let's pull that out, first:

4[x^(-3/5) + 2x^(1/5)]

We want one of these x's to end up with an exponent of 0. In our above example, we chose the smallest number. So let's do that here. Pull out x^(-3/5) and then divide each term by that. Recall that means to subtract the exponent from what we're dividing. Since it's -3/5, it's now essentially the same as adding 3/5:

4x^(-3/5)[1 + 2x^(4/5)]

You are told to use only positive exponents, so we will simplify this to resolve that negative exponent now that it's a factor of its own, by turning it into the denominator of a fraction so the final answer is:

4[1 + 2x^(4/5)] / x^(3/5)

What is your question?

4x^(-⅗) + 8x^(⅕) =

// Negative exponents in the numerator
// are equivalent to a positive exponent
// in the denominator:
// for example: a⁻² = 1/a²

4
------- + 8x^(⅕) =
x^(⅗)

// Get a common denominator:

4 + 8x^(⅕)x^(⅗)
---------------------- =
........x^(⅗)

// And simplify: Note that when you multiply
// 2 quantities with the same base, you ADD
// the exponents.
// for example: a²a³ = a²⁺³ = a⁵

4 + 8x^(⅕+⅗)
-------------------- =
........x^(⅗)

4 + 8x^(⅘)
---------------..................ANS
x^(⅗)

4x^(-⅗) + 8x^(⅕)

4x^(-⅗) = 4x^(⅕) / x^(⅘) = 4x^(⅕)·x^(-⅘), so one way to factor this---there are many correct ways---is
x^(⅕)·(4x^(-⅘) + 8).
Using positive exponents, this is
x^(⅕)·(4/x^(⅘) + 8), or even
4x^(⅕)·(1/x^(⅘) + 2).