@mathmate

- pooja195

@mathmate

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- pooja195

@mathmate

- pooja195

factor 25-y^2

- pooja195

I would just square root both right? :P

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## More answers

- mathmate

difference of two squares!

- pooja195

(x+5)(x-5) ?

- mathmate

Yep!

- pooja195

Thanks :)

- mathmate

Now,
Simplify \(\Large \frac{3x^3}{6x^2}\)

- pooja195

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

- mathmate

What you should do \(whenever\) you divide out anything from the denominator is to make sure the factor cannot be zero.
3 is not zero, so nothing needs to be done! But...

- pooja195

-_-

- mathmate

You would specify that the answer is \(\frac{x}{2}\) for x\(\ne\)0

- mathmate

is that ok?

- mathmate

Remember whenever we cancel a factor from the denominator, we are not allowed to cancel unless the factor is \(not\) equal to zero.

- mathmate

Most teachers will take half the points away if you forgot that, or some of these conditions.

- pooja195

O_O

- mathmate

That was over copied!
Simplify \(\Large \frac{x(x^2+6)}{x^2}\)

- pooja195

(x^2+6) / x

- mathmate

Condition?

- pooja195

\[x \neq1\]

- mathmate

You cancelled x, so the condition is x\(\ne\)0

- mathmate

It's what you cancelled should not equal to zero.

- mathmate

x\(\ne\)1 is for cancelling (x-1)

- mathmate

are we good?

- pooja195

Yes

- mathmate

ok, ready for the next one!

- pooja195

mhm

- mathmate

Simplify \(\Large \frac{p^2-2p+1}{p^2-1}\)

- mathmate

Remember to specify the conditions if you cancelled any factor.

- pooja195

\[\frac{ (p−1)(p−1) }{ p^2-1}\]

- mathmate

Continue! There are factors for the denominator.

- mathmate

diff of two squares!

- mathmate

In the worst case, use the quadratic formula!

- pooja195

\[\frac{ (p-1)(p-1) }{ (p+1)(p-1) }\]

- mathmate

Excellent, you can now finish off. Tag on the conditions whenever you cancel!

- pooja195

\[\frac{ p-1 }{ p+1 }~~~~~~~ x \neq 1\]

- pooja195

** -1

- mathmate

x\(\ne\)1 is correct, because it is the same as (x-1)\(\ne\)0

- mathmate

is that clear how we got x\(\ne\)1 ?

- pooja195

yes

- mathmate

ready for another?

- pooja195

no more of these ;-;

- mathmate

This is going to be all of these for the rest of 11!
Just one more of these simple ones.
Once you can do the simple ones, you'll even like the more complicated ones! lol
Simplify \(\Large \frac{3(4-m)}{6(m-4)}\)

- pooja195

2

- mathmate

you get the idea, but a little too fast ! :(

- mathmate

Remember m-4 = -(4-m), and 3/6=1/2

- mathmate

... and the conditions!

- pooja195

x /= 1/3 , 3/6

- mathmate

Well, we have
\(\Large \frac{3(4-m)}{6(m-4)}\)
= \(\Large \frac{-3(m-4)}{6(m-4)}\)
=\(\Large \frac{-3(m-4)}{3*2(m-4)}\) so we cancel 3 and (m-4)
but we know that 3\(\ne\)0, so we don't have to specify as a condition, the other one is
m-4\(\ne\)0. If we add 4 to each side, we have m\(\ne\)4

- mathmate

So the final answer is
-1/2, where m\(\ne\)4

- mathmate

are we good, especially with conditions?
We'll be working with conditions and factorization throughout ch. 11

- pooja195

yessss

- mathmate

34.
simplify \(\Large \frac{x^2-9}{x^2-5x-6}\)

- pooja195

\[\frac{ (x+3)(x-3) }{ (x-6)(x+1)}\]

- mathmate

anything else to do or to write?

- mathmate

(note: the factoring is correct! Well-done!)

- pooja195

nope thats all : P

- mathmate

If you did not cancel anything, then you make a note: the expression is already in its simplest form. (and no conditions if nothing cancelled)

- pooja195

ok :)

- mathmate

Simplify if possible \(\Large \frac{3x-5}{25-30x+9x^2}\)

- pooja195

i need help ;-;

- mathmate

Remember steps for factoring:
1. take out GCF (common factors)
2. Check if it is difference of 2 squares (Must be a difference, and only has two terms)
3. Check if it is a perfect square:
First and last terms are perfect squares, and middle term is twice the product of the square-roots of the two end terms.
Example: a^2+2ab+b^2 : End terms are perfect squares, middle = 2(a)(b), so yes, this is a perfect square, equal to (a+b)^2.
4. Try factoring.

- mathmate

1. any common factors (in the denominator)?

- pooja195

no

- mathmate

2. Is it diff. of 2 squares?

- pooja195

no

- mathmate

3. Is it a perfect square?

- pooja195

no

- mathmate

Hmm, let's see.
Square-root of first term = sqrt(25) =5
square-root of last term =sqrt(9x^2)= 3x
(so far so good, both terms are perfect squares)

- mathmate

Now do a foil on (5-3x) and see if the middle term fits.
(5-3x)(5-3x)=5^2-15x-15x+9x^2=5^2-30x+9x^2 Yay, so it's a perfect, equal to
(5-3x)^2

- mathmate

So now can you complete it?

- mathmate

Shall we go back to perfect squares?

- pooja195

\[\frac{ 1 }{ 3x-5 }\]

- mathmate

Watch out: 5-3x = -1 (3x-5) and also since we cancelled 3x-5, we specify the condition x\(\ne\)5/3

- pooja195

>.<

- mathmate

So the answer is:

- pooja195

1/3x-5
x/= 5/3

- mathmate

Yep, that's correct!

- mathmate

For perfect squares, it has to satisfy 3 conditions
1. end terms are perfect squares, i.e. you can find the square roots without leaving the radical.
2. middle term equals twice the square-roots of the end terms.
3. the middle term has the same sign of the factored terms.

- mathmate

We check that with an example
Factor 81x^2- xy +64y^2.

- mathmate

Note that condition one means that both end terms must be positive.
Here
first term = 81x^2, square root = 9x

- mathmate

Last term = 64 y^2, square-root = ?

- mathmate

@pooja195 What is the square-root of the last term?

- pooja195

3 :/ ?

- mathmate

Last term = 64 y^2, square-root = ?

- pooja195

8y

- mathmate

Good.
What's twice the product of the square-roots?

- pooja195

huh ?

- mathmate

We need to multiply together the square-roots (of the first and third terms), and double it.

- pooja195

512? :/

- pooja195

im so confused :/

- mathmate

the square roots are 9x and 8y,so the product is 72xy, and doubling it shoud give 144xy.
This should match the middle term (if the expression is a perfect square).
It doesn't because I forgot to fill in the 144 in the middle.
So we conclude that (9x+8y)^2=81x^2+144xy+64y^2=(9x+8y)^2.

- mathmate

|dw:1433718072207:dw|

- mathmate

|dw:1433718524311:dw|

- mathmate

So if we are given 81x^2+144xy+64y^2
We can do the reverse of FOIL, assuming that it is a perfect square.
We find 9x by sqrt, and 8y by sqrt.
Then we test if 2(9x)(8y)=144xy, if it's equal (9x+8y)^2 are the factors.

- mathmate

|dw:1433718696561:dw|

- pooja195

Thats so much more better :)

- pooja195

I understand this a bit more :o

- mathmate

Good! An image is worth 1000 words, I was told!

- pooja195

xD

- mathmate

So we'll be back on track!?

- pooja195

yes

- mathmate

11.4 Multiplying and dividing rational expresssions
===================================================
calculate \(\Large \frac{6}{15}\times \frac{12}{9}\)

- mathmate

I would like you to treat the numbers as products of factors, and see if you can simplify without the calculator.

- pooja195

i cant do it without a calc! :O

- mathmate

It's not the answer we want, just the process. I'll do that for you.

- mathmate

\Large \frac{2.3}{3.5}\times \frac{2.2.3}{3.3}

- mathmate

\(\Large \frac{6}{15}\times \frac{12}{9}\)
=\(\Large \frac{2.3}{3.5}\times \frac{2.2.3}{3.3}\)
now cancel factors
=\(\Large \frac{2}{5}\times \frac{2.2}{3}\)
=\(\Large \frac{8}{15}\)

- mathmate

If you replace each prime factor (2,3,5,...) by a polynomial, this is exactly what we do to multiply rational expressions!

- pooja195

hmm ok

- mathmate

simplify
\(\Large \frac{x}{3x^2-9x}\times\frac{x-3}{2x^2+x-3}\)
[do not forget the conditions whenever you cancel]

- pooja195

\[\frac{ (x-3) }{ 3(x−1)(2x+3)(x−3) }\]

- pooja195

correct so far?

- mathmate

exactly, continue!

- pooja195

\[\frac{ 1 }{ 3(x-1)(2x+3) }\]
x/= 3

- mathmate

Almost.
The last step is all correct.
There was a mistake in the first step:

- mathmate

Oh, actually, you were right, I was wrong.
The final answer is good! Congrats!

- pooja195

O_o

- mathmate

Do you still remember how to do a division of fractions?

- mathmate

I'd say: flip the fraction after the \(\div\) sign.

- mathmate

Do you agree?

- pooja195

Can we keep doing multiplication ?

- mathmate

Sure! Another one coming up

- mathmate

\(\Large \frac{c^2-64}{4c^3} \times \frac{c}{c^2+9c+8}\)

- mathmate

Simplify!

- pooja195

\[\frac{ (c+8)(c−8) }{ 4cc(c+1)(c+8) }\]

- pooja195

correct so faR?

- mathmate

Excellent, go on!

- mathmate

Remember you cancelled a "c" already. Keep track of that.

- pooja195

Can you finish it off ;-;

- mathmate

\(\Large \frac{c^2-64}{4c^3} \times \frac{c}{c^2+9c+8}\)
=\(\Large \frac{(c+8)(c-8)}{4c^3} \times \frac{c}{(c+8)(c+1)}\) cancel c and (c+8)
=\(\Large \frac{(c-8)}{4c^2(c+1)}~~~ x\ne -8,0\)

- mathmate

Is the last part ok for you now?

- pooja195

yes :)

- mathmate

Another multiplication?

- pooja195

nope ;-;

- mathmate

Division:
\(\Large \frac{1}{2}\div\frac{1}{5}\)
=\(\Large \frac{1}{2}\times\frac{5}{1}\)
Is this familiar to you?

- pooja195

yes

- mathmate

We would do the same with rational functions.

- mathmate

Gimme a second to type it up.

- mathmate

simplify
\(\Large \frac{x}{x+6} \div \frac{x+3}{x^2-36}\)

- pooja195

\[\frac{ x }{ x+6 }\times \frac{ (x+6)(x-6) }{( x+3) }\]
\[\frac{ x(x-6) }{ (x+3) }~~~~~x \neq6\]

- mathmate

Perfect!!!! Can't be better!

- pooja195

REALLY? :O

- pooja195

I did that without a calc :O

- mathmate

Oops, what about the previous ones ! LOL

- pooja195

smh

- mathmate

Ready for another?

- pooja195

yes!: )

- mathmate

Simplify
\(\Large \frac{45x^3-9x^2}{x} \div \frac{6(x-5)}{2}\)

- pooja195

\[\frac{ 3x(5x−1) }{ x-5 }\]

- mathmate

Wow, what did you eat for supper?

- pooja195

xD

- mathmate

Another?

- pooja195

No i have this : ) next case!

- mathmate

k. coming right up!

- mathmate

11.5 Adding and Subtracting with like denominators
=================================

- mathmate

Like denominators mean that the denominators are either the same, or just a simple multiple.

- mathmate

So it's a matter of adding the numerators, like:|dw:1433721592963:dw|

- pooja195

ok

- mathmate

now gimme a minute.

- mathmate

Add
\(\Large \frac{x+2}{x} + \frac{3x-2}{x}\)

- pooja195

\[\frac{ 4x }{ x }\]

- mathmate

...and the next step is...

- pooja195

the answer would be 4

- pooja195

Because the x's cancel right?

- mathmate

Yes, ...and...

- pooja195

and??

- mathmate

x\(\ne\)0
"conditions whenever cancelling a factor in the denominator".

- pooja195

i knew it! >:(

- mathmate

haha!
Another?

- pooja195

Next case!

- mathmate

\(\Large \frac{3x-4}{x-4} - \frac{2x}{x-4}\)

- mathmate

Just curious!

- pooja195

i knew it was coming :P
\[\frac{ 1x-4 }{ x-4}\]

- mathmate

recall 1y = y

- pooja195

1

- mathmate

and..

- pooja195

and??

- mathmate

x\(\ne\)4

- mathmate

The conditions are the only difficult part in these examples!

- pooja195

ugh

- mathmate

|dw:1433722130469:dw|

- pooja195

multiplication right?

- mathmate

Nope, perimeter would be the sum of all four sides.

- pooja195

Next case!

- mathmate

Please!

- mathmate

We only have two more cases.

- pooja195

I dont like this one ;-;

- pooja195

im not even sure where to start .-.

- mathmate

|dw:1433722487080:dw|

- mathmate

can you finish it?

- pooja195

\[\huge~\frac{ 2(x−2)(x^3+4x^2−12x+4) }{ (x+2)(x−2)(x−2) }\]

- mathmate

wow, not like that.
The addition had already been done, because the denominators are the same.

- mathmate

All you need to do is to cancel (x-2) and give the answer!

- mathmate

|dw:1433722913558:dw|

- pooja195

>.<

- mathmate

Do you see what was happening?
I multiplied by 2 because the opposite sides are identical.
I then add them, using the common denominator that we are given with.
Did the additions to come to 2(x+4)(x-2)/(x-2)
Since (x-2) is a common factor, we can cancel as long as x-2\(\ne\)0, or x\(\ne\)2
So the answer is 2(x+4), with x\(\ne\)2

- pooja195

Got it

- mathmate

good!

- mathmate

another?

- pooja195

no ;;

- mathmate

ok, next case:
11.6 adding and subtracting rational functions ith unlike denominators
=============================================
It's almost the same as 11.5.
It's just you need to do dancing class with the denominator before adding and subtracting.

- mathmate

example
Find common denominator of
\(\Large \frac{1}{12x} +\frac{2+x}{40x^4}\)
The common denominator can be found similar to the dancing class:
|dw:1433723859840:dw|

- mathmate

Once you can find the common denominator readily, rest is relatively straight-forward.

- pooja195

Makes sense :)

- mathmate

So that's good, we'll proceed with an example.

- mathmate

Simplify
\(\Large \frac{x+1}{5} + \frac{2x}{6}\)
Can you first try to find the common denominator?

- mathmate

|dw:1433726621112:dw|

- mathmate

Do you agree?

- pooja195

Yess

- pooja195

\[\huge~\frac{8x+3}{15} \]

- mathmate

Exactly, so we go with the same idea for the other rational expressions.

- mathmate

Now do you still remember equivalent fractions?

- pooja195

yes

- mathmate

for example,
\(\Large \frac{9}{5x} = \frac{? }{40x^4 }\)

- pooja195

8?

- mathmate

It's basically cross multiplication,
\(\Large ? =40x^4 ~\frac{9}{5x} = 72x^3\)
Dose that make sense?

- mathmate

so
\(\Large \frac{9}{5x}= \frac{72x^3}{40x^4}\)

- mathmate

is that ok? We need to use this step later on.

- pooja195

yes its fine

- mathmate

Ok, try to find the sum in its simplest form:
\(\Large \frac{1}{x^2} + \frac{2}{x}\)

- pooja195

\[\huge~\frac{2x+1}{x^2}\]

- mathmate

Yes, that correct!

- pooja195

\[\large~\frac{ 3 }{ x }\]

- mathmate

cannot do that, because x and 1 are not like terms, so there is nothing to factor out.

- pooja195

oh right

- mathmate

Add and simplify if possible \(\Large \frac{1}{x+1} + \frac{1}{x-1}\)

- pooja195

\[\huge~\frac{2x}{\left(x+1\right)\left(x-1\right)} \]

- mathmate

Wow, are we wasting time?

- mathmate

Try this: add
\(\Large \frac{x-5}{x+5} + \frac{x+2}{x-2}\)

- mathmate

Sorry, subtract!

- mathmate

\(\Large \frac{x-5}{x+5} - \frac{x+2}{x-2}\)

- mathmate

Use FOIL.

- pooja195

\[\huge~\frac{14x}{\left(x+5\right)\left(x-2\right)}\]

- mathmate

\(\Large \frac{x}{x-10} + \frac{x+4}{x+6}\)

- pooja195

wait i cant do this one .-.

- mathmate

When there is no common factor, you only have to "kind of" cross multiply.

- pooja195

\[\frac{2x^2-40}{\left(x+6\right)\left(x-10\right)} \]
i have it uptil this part

- mathmate

|dw:1433728624180:dw|

- mathmate

Yes, it's correct, but need to factor out the 2 to give 2(x^2-20) at the numerator.

- mathmate

Another one?

- pooja195

nuu next case!

- mathmate

ok. Have you done rational functions?

- mathmate

p.678

- mathmate

11.7 Rational Functions
==============

- pooja195

Havent done that

- mathmate

Sure?

- pooja195

yes

- mathmate

ok, good.
Together with the calculator(s), we have covered everything!

- pooja195

._. lol thanks for all the help :)

- mathmate

You're welcome! :)

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