Question

(x^2+4x/x^2+4)([x+2]^2/x^2) How do I simplify this? Please help and give detailed instructions, I will give you a shiny medal :D

Answer 1

\[\frac{ (x^2+4x)}{ x^2+4 }\times \frac{ (x+2)^2}{ x^2 }\]

Answer 2

so, let's start with the easy part.

Answer 3

do \(x^2~and~4x\) share anything in common?

Answer 4

no, they aren't the same :D because x^2 is grouped with itself, so we leave 4x alone right?

Answer 5

or do I expand it out?

Answer 6

well, they both have an x right?

Answer 7

yes they do, so is it x(x+4)?

Answer 8

very nice

Answer 9

so, that means we can cancel out an x like so:

Answer 10

\[\frac{x^2+4x}{x^2+4}\times \frac{(x+2)^2}{x^2}=\frac{\not{x}(x+4)}{x^2+4}\times \frac{(x+2)^2}{x^{\not{2}}}\]

Answer 11

now, lets expand that x+2 squared

Answer 12

can you do that?

Answer 13

so I end up with: \[\frac{ x(x+4) }{ x(x)+4 }\] so x gets cancelled out? and it becomes just 4/4 = 1 and \[\frac{ (x+2)^2 }{ x^2 } becomes \frac{ (x+2) (x+2) }{ x(x) }\]

Answer 14

uhm, not quite

Answer 15

very nice try though

Answer 16

oh :( I'm sorry, what did I do wrong?

Answer 17

its hard to explain

Answer 18

but, you cannot pull the x out of only one term
although you wrote it correctly

Answer 19

your expansion of the (x+2)(x+2) was just what we need though

Answer 20

oh, so I leave the x(x+4) and x^2+4 alone for now then?

Answer 21

exactly so we are here: \[\frac{x^2+4x}{x^2+4}\times \frac{(x+2)^2}{x^2}=\frac{\not{x}(x+4)}{x^2+4}\times \frac{(x+2)^2}{x^{\not{2}}}\\ =\frac{x+4}{x^2+4}\times \frac{(x+2)(x+2)}{x}\]

Answer 22

What we want to do now is factor the \(x^2+4\)

Answer 23

so is it: x(x) + 4?

Answer 24

ok, so technically that is a correct statement, but not what we are looking for. Have you ever heard of a conjugate?

Answer 25

no not really, :( I heard it but my teacher didn't really explain it in depth..

Answer 26

does it become negative then?

Answer 27

it's ok, I just noticed it was a plus not a minus. let me think for a moment. If you want to know what it is in the mean time heres a reference.(i was trying to use tricks instead of synthetic division) http://www.purplemath.com/modules/radicals4.htm

Answer 28

ahh oki, thanks for the link, I'll read it :D

Answer 29

This is simply a multiplication of fractions.
Factor each numerator and denominator.
Multiply the numerators together and multiply the denominators together.
There is only very little reducing you can do.

Answer 30

So, I'm thinking ethos point we do all of the multiplication then use synthetic division to get it as simple as possible

Answer 31

\(\dfrac{ x^2+4x}{ x^2+4 }\times \dfrac{ (x+2)^2}{ x^2 }=\)

\(=\dfrac{ x(x+4)}{ x^2+4 }\times \dfrac{ (x+2)^2}{ x^2 }\)

\(=\dfrac{ x(x+4) (x+2)^2}{x^2 (x^2+4) }\)

\(=\dfrac{ (x+4) (x+2)^2}{x (x^2+4) }\)

Answer 32

*at this point*

Answer 33

so the outside x on the 4 got removed? and cancelled out the x2(x2+4)

Answer 34

I mean x^2(x^2+4)

Answer 35

Explanation for above:
Line 1. The given problem.
Line 2. Factor the left numerator.
Line 3. Multiply the numerators together and multiply the denominators together.
Line 4. Reduce. \(\dfrac{x^2}{x} = \dfrac{x}{1} \)

Answer 36

yea, just the x and x^2 canceled at the current step

Answer 37

so, anyways when we reduce this the rest of the way, we will need synthetic division. I performed the multiplication math student and I had set up.

Answer 38

\[\frac{x^3+8x^2+20x+16}{x^3+4x}\]

Answer 39

If you have not learned synthetic division, I would accept this as the most simplified version for my classes; however, if you have covered synthetic division we can continue

Answer 40

I'd also accept if the top had not been multiple out too

Answer 41

so you got that by doing (x+2)(x+2) = x^2+4x^2+4x+4x^2+16x+16

Answer 42

how did it become x^3?

Answer 43

(x+4)(x+2)(x+2)

Answer 44

there are 3 terms

Answer 45

so it was x(x^2+4) so x(x^2) = x^3 and 4(x) = 4x ? :) :o

Answer 46

very good, that is exactly how i git the denominator

Answer 47

got*

Answer 48

so now, have you covered synthetic division?

Answer 49

If you need to see it to know, here's a reference http://www.purplemath.com/modules/synthdiv.htm

Answer 50

\[\frac{x^3+8x^2+20x+16x^3+4x }{ 1 } \div \frac{ x^3+4x}{ 1 }\]

Answer 51

oh oops

Answer 52

so, that is the same as a single fraction.

Answer 53

where are you trying to head with that?

Answer 54

i think the fraction is probably all they are looking for at this point

Answer 55

flip over the 4x^3 and get the denominator the same and multiply across but I should try the synthetic division

Answer 56

actually, polynomial will be faster

Answer 57

use the first method

Answer 58

oh ok, so (x^3 +4x) (?) = (x^3+8x?) is that the first step?

Answer 59

I multiplied by 2 first so I could get x^3+8x - 2x^3+8x

Answer 60

Mathematica's "Apart" function yields the following:\[\frac{4 (x+4)}{x^2+4}+\frac{4}{x}+1 \]

Answer 61

oh, so is that the answer?

Answer 62

The answer is yes if it looks more simple then the original.

Answer 63

thats not an answer choice though :<

Answer 64

so i multiplied it by 1 so x^3+4x would be: