Question

@mathmate

Answer 1

Yep, you have a question? What's the topic?

Answer 2

Things like this :/
http://prntscr.com/79g79q

Answer 3

Oh, rational functions!

Answer 4

When you have to simplify a rational function, first thing you have to remember:
"the denominator cannot be zero.", else the expression is not defined (infinite).

Answer 5

Next, to simplify it, both must have common factors.
So we need to factorize as far as possible both numerator and denominator.
So far so good?

Answer 6

Can you do that for me (the factorization)?

Answer 7

umm
\[\frac{ 3\times~x~\times~x }{ 3x(2x-y) }\]

Answer 8

Perfecct!

Answer 9

The next step is to simplify by cancelling common factors between top (numerator) and bottom (denominator).

Answer 10

Remember that we are NOT allowed to cancel zeroes.
For example, 5*0=6*0, if we cancel zeroes, then we get 5=6 , which is absurd.

Answer 11

Since we don't know if x is zero, so we cancel x (top and bottom) by stating the condition x\(\ne0\).
Can you do that for me ? (cancel common factors, and if a factor could be zero, state the condition).

Answer 12

You can cancel 3 with 3 without stating anything, because 3 is not zero.

Answer 13

so...
\[\frac{ x }{ (2x-y)}\]
is that right? ;-;

Answer 14

That is almost correct. Perfect if you had stated the condition \(x\ne0\)

Answer 15

like:
\(\large \frac{ x }{ (2x-y)}\) where x\(\ne0\)

Answer 16

hmmm ok :/

Answer 17

To sum it up: any factor that you cancel, you have to add the condition ( ) \(\ne\) 0.
If you cancelled 3 different factors, then you would state ( ), ( ), ( )\(\ne\)0.
Non-zero numerical factors don't count, because you already know they are not zero.

Answer 18

Is that ok so far before you post another example (of the same topic)?

Answer 19

Yea thats fine i understand that part now :) and I want u to post something(the problem) so i dont end up cheating T_T

Answer 20

ok, coming up.

Answer 21

simplify
\(\large \frac{6x^2+7x-3}{6x^2-11x+3}\)

Answer 22

@kidrah69 Do show the intermediate steps, and don't bother with tex format if it's easier for you.

Answer 23

soo many x's o.o
\[\frac{ 6*x*x*7*x*x*x*x*x*x*x*-3 }{ }\]
is this right far T_T

Answer 24

Not really. Remember how to factorize a quadratic expression, like:
\(x^2+3x+2=(x+1)(x+2)\)
Rational function simplification requires factorization as a tool!

Answer 25

It would have been 6∗x∗x+7∗x−3
but won't help here because x is not a common factor. You need to factorize the quadratic expressions using what you learned before.

Answer 26

Do you want me to do the top (numerator)?

Answer 27

\(6x^2+7x-3\)
m*n = -18
m+n=7
So we are looking for m, n such that the sum is 7, and the product -18 (m, n have different signs, one positive, one negative). The positive number is bigger (so the sum is positive)
m n sum
-1,18 17
-2,9 7 ok that's what we want
so
\(6x^2+7x-3=6x^2-2x + 9x-3=2x(3x-1) + 3(3x-1)=(3x-1)(2x+3)\)

Answer 28

I'll give you some time to finish the denominator, and we'll try a simpler one.
Say
Simplify \(\large \frac{5x^2+3x}{2x^2}\)

Answer 29

\[\frac{ 5*x*x+3*x*x *3}{2*x*x }\]

Answer 30

This is similar to the first one which you did very well.
The expanded form should look like:
\(\large \frac{ 5*x*x+3*x }{2*x*x }\)

Answer 31

\[\frac{ 15x }{ 2 }\]

Answer 32

Recall that there is a plus sign, so you cannot multiply 5 and 3.
The two terms 5x^2 and 3x are not like terms because they do not have the same power of x.
However, you can factorize the x out, to give
5x^2+3x = x(5x+3)
and the bottom you can write
x(2x)
to make
\(\large \frac{5x^2+3x}{2x^2}=\frac{x(5x+3)}{x(2x)}=\frac{5x+3}{2x}\) where \(x\ne0\)

Answer 33

oh right >_< my bad ok that makes more sense :)

Answer 34

Is that ok so far?

Answer 35

Yes.

Answer 36

Try another?

Answer 37

Can we work on this later?

Answer 38

sure! do you have another question?

Answer 39

As of now know i still need to go back and look at what im not comfortable with >_<

Answer 40

*no

Answer 41

Yes, reviewing factorization would help you in the rational functions!
math is like that. Like you're stepping on a chair to get something from the top shelves.
If the chair is not solid, you would fall!

Answer 42

Right :)

Answer 43

So whenever you finish your review and ready for some other questions, tag me.

Answer 44

haha ok :D thanks

Answer 45

@mathmate

Answer 46

yep

Answer 47

We'll see if this is better!

Answer 48

Hi, you have a question?

Answer 49

Yes, this browser seems to work better!

Answer 50

Ohhh :DDD im not sure :/ can we continue with yesterday's work? :)

Answer 51

Yes, tell me from where!

Answer 52

Are you referring to the factoring problem?

Answer 53

yep :-)

Answer 54

There are two kinds of factoring, the simple one, such as:
x^2+2x = x(x+2)
This means there is a single factor (x) in each and every one of the terms.
Then there is quadratic factoring, such as:

Answer 55

\(6x^2+7x-3\)
where you need to use the m,n rules to do.
Sometimes when you get lucky, there are other patterns (perfect squares and difference of two squares)

Answer 56

Would you like me to do the \(6x^2+7x-3\) again from scratch?

Answer 57

right so far?

Answer 58

i cant think of factors :/

Answer 59

You mean:

Answer 60

Well, since -18 is negative, so the two numbers (m, n) must have different signs, i.e. one positive, and the other is negative. Agree?

Answer 61

Yes

Answer 62

So we can list the factors with first one positive, and the other negative. If the sum comes up to be -7 (instead of 7, we just reverse the signs. ok so far?

Answer 63

Can you continue the table like this:
m n sum
-1 18 17
...

Answer 64

We look for a sum of +7

Answer 65

:/ im confused on finding the sum .-.

Answer 66

R U ok with the factors -1 and 18?

Answer 67

The sum is -1+18=18-1=17

Answer 68

Right but thats not 7 :/

Answer 69

No, that's why you need to continue with -2 and 9 (with a product of -18)

Answer 70

... and so on.

Answer 71

until you get 7.

Answer 72

Its 9 and -2 xD

Answer 73

Yep! (you're lucky this time :)
So what's the next step?

Answer 74

knowing that 9x-2x = 7x

Answer 75

You would replace 7x by 9x-2x and continue, like:
\(6x^2+7x-3=6x^2+9x -2x-3\) and factor the first and second groups.

Answer 76

6x^2 -2x + 9x -3
2x(x-1)+ 3(x-1)
(2x+3)
(x-1)

Answer 77

almost...
2x(3x-1)+ 3(3x-1)
so
(2x+3)(3x-1)

Answer 78

Very good! Want to try another one?

Answer 79

Sure. :)

Answer 80

Factor \(x^2+2x-15\).

Answer 81

(x+5)(x-3)

Answer 82

Excellent, can you show your work?