Simplifying Equations with Polynomial denominators/numerators

[darshanpatel]

Homework Statement

I uploaded a picture of the question because I didn't want to confuse people from the way it looks because it is pretty long.

http://tinypic.com/r/2itpm39/5

Homework Equations

-None-

The Attempt at a Solution

I went from the original equation and took an 'x' out of the first numerator and then got the reciprocal of the x^n and multiplied that through

1. Original Equation
2. (((x^n)-8)/((x^2n)+(2x^n)+1)) x ((x^2n)-(4x^n)-5)/(x^n)

I don't know what I can factor out from this point..
Please help and thanks

 

[SammyS]

Homework Statement

I uploaded a picture of the question because I didn't want to confuse people from the way it looks because it is pretty long.

http://tinypic.com/r/2itpm39/5

Homework Equations

-None-

The Attempt at a Solution

I went from the original equation and took an 'x' out of the first numerator and then got the reciprocal of the x^n and multiplied that through

1. Original Equation
2. (((x^n)-8)/((x^2n)+(2x^n)+1)) x ((x^2n)-(4x^n)-5)/(x^n)

I don't know what I can factor out from this point..
Please help and thanks

x²ⁿ = (xⁿ)²

That in itself may help you. If not, then let u = xⁿ and substitute that.

 

[darshanpatel]

Do you mean x^2n= (x^2)^n

Also as an example, if I took x^n-8 and did that, would it be:

((x)^3n)-(2)^3
and i don't know where to go from there

and for this one:

(x^2n)-(4x^n)-5
((x^2)^n)-((4x)^n)-5
what would i do next?

 

[eumyang]

Do you mean x^2n= (x^2)^n

While that's true, SammyS was right to say
x²ⁿ = (xⁿ)²

So if you were to make the substitution u = xⁿ,
xⁿ - 8 would become u - 8.

and for this one:

(x^2n)-(4x^n)-5
((x^2)^n)-((4x)^n)-5
what would i do next?

With the substitution, it would be
x²ⁿ - 4xⁿ - 5
= (xⁿ)² - 4xⁿ - 5
= u² - 4u - 5

Now do the same kind of substitution in the first denominator. Doesn't that expression look familiar now? And can't you do something with u² - 4u - 5?

 

[darshanpatel]

I substituted all the x^n with y's and ended up with:

(((x^n)-5)((x^n)-2)((x^2n)+(2x^n)+4))/(x^n)((x^n)+1)

Is that all it can get factored and simplified?

 

[darshanpatel]

Also, I think i will lose points by substituting, what is the way without it?

 

[SammyS]

I substituted all the x^n with y's and ended up with:

(((x^n)-5)((x^n)-2)((x^2n)+(2x^n)+4))/(x^n)((x^n)+1)

Is that all it can get factored and simplified?

((xⁿ ) - 2) ((x²ⁿ ) + (2xⁿ ) + 4) = x³ⁿ - 8 ≠ xⁿ - 8

 

[darshanpatel]

I redid it and got: (((x^n)-8)((x^n)-5))/(x^n)((x^n)+1)

Is that correct?

Sorry, I didn't get what you are trying to say

 

[eumyang]

I redid it and got: (((x^n)-8)((x^n)-5))/(x^n)((x^n)+1)

Is that correct?

If you mean
$$\frac{(x^n-8)(x^n-5)}{x^n(x^n+1)}$$
... then yes, looks right.

 

[Mentallic]

Also, I think i will lose points by substituting, what is the way without it?

Substituting isn't necessary, but makes things look simpler.

For example, if you were asked to factorize

$$x^2+4ax+4a^2$$

then it isn't very clear what needs to be done, but if you let 2a=b then you have

$$x^2+2bx+b^2$$ and this is clearly a perfect square, so you can factorize it as so:

$$(x+b)^2$$ and then you can substitute back in at the end to obtain

$$(x+2a)^2$$