(x^2-2x+4/x^2-5x+6)divided by(x^3+8/x^2-9)

- anonymous

(x^2-2x+4/x^2-5x+6)divided by(x^3+8/x^2-9)

- schrodinger

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- anonymous

\[[(x^2-2x+4)\div(x^2-5x+6)] \div[(x^3+8/x^2-9)]\]

- anonymous

Multiply the second part by the numerator and simplify.

- anonymous

I factored down to the following:
\[[(x-2)(x-2)/(x-6)(x-1)]*[(x-3)(x+3)/(x+2)(x^2-2x+4)]\]

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## More answers

- anonymous

But there are no like terms, so do you just combine all of the factors into one problem?
I got this as my answer so far:
\[(x-2)^2(x-3)(x+3)/(x-1)(x-6)(x+2)(x^2-2x+4)\]

- anonymous

You flip the second part, then multiply them out. Factoring is unnecessary.

- anonymous

I did flip the terms, but the bases aren't the same.

- anonymous

You're multiplying. It doesn't matter if the bases are the same. That only matters with addition and subtraction of fractions.

- anonymous

So my answer is still technically right, I just have to multiply everything out?

- anonymous

Yes, but I wouldn't.

- anonymous

Leave it unfactored, flip it, and multiply through.

- anonymous

But why include fractions that contain the difference and sums of two squares and cubes if you weren't supposed to factor them?

- anonymous

Because you're not solving for the roots of x.
You'd have to multiply it all out all over again anyway, and make things VERY difficult for yourself. Itd be redundant.

- anonymous

Multiplying all of that out would take an entire page, and isn't factoring simplifying a problem? My instructions say to simplify completely.

- anonymous

(didn't include that above, which is my bad)

- anonymous

You have to factor AFTER simplifying, and AFTER multiplying... Otherwise nothing is simplified, and it's redundant and unnecessary.

- anonymous

That and you didn't factor correctly, so I'd want to avoid that.

- anonymous

So you are saying I should take \[(x^2-2x+4)(x^2-9)\] and so on?

- anonymous

Yes. Then post what you get.

- anonymous

Oh...

- anonymous

The numerator should be x^4 -2x^3 + 4x^2 -9x^2 +18x -36
simplify and factor that

- anonymous

\[x^4-2x^3-5x^2-18x-36\] for above then?

- anonymous

I had -9x^2+4x^2=-5x^2. Does that check out with you?

- anonymous

Numerator yes. I'm not sure what you mean by the second post though

- anonymous

For the numerator, it is fine though. That gives me plenty to work with, thank you again :D

- anonymous

\[\left(\frac{4-2 x+x^2}{(-3+x) (-2+x)}\right)\left(\frac{(-3+x) (3+x)}{(2+x) \left(4-2 x+x^2\right)}\right)=\frac{3+x}{-4+x^2} \]

- anonymous

Left Side = Right Side when both are evaluated with x=19

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