Question

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

Answer 1

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

Answer 2

Multiply the second part by the numerator and simplify.

Answer 3

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

Answer 4

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:
\[(x-2)^2(x-3)(x+3)/(x-1)(x-6)(x+2)(x^2-2x+4)\]

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

Yes, but I wouldn't.

Answer 10

Leave it unfactored, flip it, and multiply through.

Answer 11

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

Answer 12

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.

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

Yes. Then post what you get.

Answer 19

Oh...

Answer 20

The numerator should be x^4 -2x^3 + 4x^2 -9x^2 +18x -36

simplify and factor that

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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