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factor 25-y^2
I would just square root both right? :P

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difference of two squares!
(x+5)(x-5) ?
Yep!
Thanks :)
Now, Simplify \(\Large \frac{3x^3}{6x^2}\)
\[\frac{ 3*x*x*x }{ 3*2*x*x }\] \[\frac{ x }{ 2 }\]
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...
-_-
You would specify that the answer is \(\frac{x}{2}\) for x\(\ne\)0
is that ok?
Remember whenever we cancel a factor from the denominator, we are not allowed to cancel unless the factor is \(not\) equal to zero.
Most teachers will take half the points away if you forgot that, or some of these conditions.
O_O
That was over copied! Simplify \(\Large \frac{x(x^2+6)}{x^2}\)
(x^2+6) / x
Condition?
\[x \neq1\]
You cancelled x, so the condition is x\(\ne\)0
It's what you cancelled should not equal to zero.
x\(\ne\)1 is for cancelling (x-1)
are we good?
Yes
ok, ready for the next one!
mhm
Simplify \(\Large \frac{p^2-2p+1}{p^2-1}\)
Remember to specify the conditions if you cancelled any factor.
\[\frac{ (p−1)(p−1) }{ p^2-1}\]
Continue! There are factors for the denominator.
diff of two squares!
In the worst case, use the quadratic formula!
\[\frac{ (p-1)(p-1) }{ (p+1)(p-1) }\]
Excellent, you can now finish off. Tag on the conditions whenever you cancel!
\[\frac{ p-1 }{ p+1 }~~~~~~~ x \neq 1\]
** -1
x\(\ne\)1 is correct, because it is the same as (x-1)\(\ne\)0
is that clear how we got x\(\ne\)1 ?
yes
ready for another?
no more of these ;-;
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 \(\Large \frac{3(4-m)}{6(m-4)}\)
2
you get the idea, but a little too fast ! :(
Remember m-4 = -(4-m), and 3/6=1/2
... and the conditions!
x /= 1/3 , 3/6
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
So the final answer is -1/2, where m\(\ne\)4
are we good, especially with conditions? We'll be working with conditions and factorization throughout ch. 11
yessss
34. simplify \(\Large \frac{x^2-9}{x^2-5x-6}\)
\[\frac{ (x+3)(x-3) }{ (x-6)(x+1)}\]
anything else to do or to write?
(note: the factoring is correct! Well-done!)
nope thats all : P
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)
ok :)
Simplify if possible \(\Large \frac{3x-5}{25-30x+9x^2}\)
i need help ;-;
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.
1. any common factors (in the denominator)?
no
2. Is it diff. of 2 squares?
no
3. Is it a perfect square?
no
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)
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
So now can you complete it?
Shall we go back to perfect squares?
\[\frac{ 1 }{ 3x-5 }\]
Watch out: 5-3x = -1 (3x-5) and also since we cancelled 3x-5, we specify the condition x\(\ne\)5/3
>.<
So the answer is:
1/3x-5 x/= 5/3
Yep, that's correct!
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.
We check that with an example Factor 81x^2- xy +64y^2.
Note that condition one means that both end terms must be positive. Here first term = 81x^2, square root = 9x
Last term = 64 y^2, square-root = ?
@pooja195 What is the square-root of the last term?
3 :/ ?
Last term = 64 y^2, square-root = ?
8y
Good. What's twice the product of the square-roots?
huh ?
We need to multiply together the square-roots (of the first and third terms), and double it.
512? :/
im so confused :/
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.
|dw:1433718072207:dw|
|dw:1433718524311:dw|
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.
|dw:1433718696561:dw|
Thats so much more better :)
I understand this a bit more :o
Good! An image is worth 1000 words, I was told!
xD
So we'll be back on track!?
yes
11.4 Multiplying and dividing rational expresssions =================================================== calculate \(\Large \frac{6}{15}\times \frac{12}{9}\)
I would like you to treat the numbers as products of factors, and see if you can simplify without the calculator.
i cant do it without a calc! :O
It's not the answer we want, just the process. I'll do that for you.
\Large \frac{2.3}{3.5}\times \frac{2.2.3}{3.3}
\(\Large \frac{6}{15}\times \frac{12}{9}\) =\(\Large \frac{2.3}{3.5}\times \frac{2.2.3}{3.3}\) now cancel factors =\(\Large \frac{2}{5}\times \frac{2.2}{3}\) =\(\Large \frac{8}{15}\)
If you replace each prime factor (2,3,5,...) by a polynomial, this is exactly what we do to multiply rational expressions!
hmm ok
simplify \(\Large \frac{x}{3x^2-9x}\times\frac{x-3}{2x^2+x-3}\) [do not forget the conditions whenever you cancel]
\[\frac{ (x-3) }{ 3(x−1)(2x+3)(x−3) }\]
correct so far?
exactly, continue!
\[\frac{ 1 }{ 3(x-1)(2x+3) }\] x/= 3
Almost. The last step is all correct. There was a mistake in the first step:
Oh, actually, you were right, I was wrong. The final answer is good! Congrats!
O_o
Do you still remember how to do a division of fractions?
I'd say: flip the fraction after the \(\div\) sign.
Do you agree?
Can we keep doing multiplication ?
Sure! Another one coming up
\(\Large \frac{c^2-64}{4c^3} \times \frac{c}{c^2+9c+8}\)
Simplify!
\[\frac{ (c+8)(c−8) }{ 4cc(c+1)(c+8) }\]
correct so faR?
Excellent, go on!
Remember you cancelled a "c" already. Keep track of that.
Can you finish it off ;-;
\(\Large \frac{c^2-64}{4c^3} \times \frac{c}{c^2+9c+8}\) =\(\Large \frac{(c+8)(c-8)}{4c^3} \times \frac{c}{(c+8)(c+1)}\) cancel c and (c+8) =\(\Large \frac{(c-8)}{4c^2(c+1)}~~~ x\ne -8,0\)
Is the last part ok for you now?
yes :)
Another multiplication?
nope ;-;
Division: \(\Large \frac{1}{2}\div\frac{1}{5}\) =\(\Large \frac{1}{2}\times\frac{5}{1}\) Is this familiar to you?
yes
We would do the same with rational functions.
Gimme a second to type it up.
simplify \(\Large \frac{x}{x+6} \div \frac{x+3}{x^2-36}\)
\[\frac{ x }{ x+6 }\times \frac{ (x+6)(x-6) }{( x+3) }\] \[\frac{ x(x-6) }{ (x+3) }~~~~~x \neq6\]
Perfect!!!! Can't be better!
REALLY? :O
I did that without a calc :O
Oops, what about the previous ones ! LOL
smh
Ready for another?
yes!: )
Simplify \(\Large \frac{45x^3-9x^2}{x} \div \frac{6(x-5)}{2}\)
\[\frac{ 3x(5x−1) }{ x-5 }\]
Wow, what did you eat for supper?
xD
Another?
No i have this : ) next case!
k. coming right up!
11.5 Adding and Subtracting with like denominators =================================
Like denominators mean that the denominators are either the same, or just a simple multiple.
So it's a matter of adding the numerators, like:|dw:1433721592963:dw|
ok
now gimme a minute.
Add \(\Large \frac{x+2}{x} + \frac{3x-2}{x}\)
\[\frac{ 4x }{ x }\]
...and the next step is...
the answer would be 4
Because the x's cancel right?
Yes, ...and...
and??
x\(\ne\)0 "conditions whenever cancelling a factor in the denominator".
i knew it! >:(
haha! Another?
Next case!
\(\Large \frac{3x-4}{x-4} - \frac{2x}{x-4}\)
Just curious!
i knew it was coming :P \[\frac{ 1x-4 }{ x-4}\]
recall 1y = y
1
and..
and??
x\(\ne\)4
The conditions are the only difficult part in these examples!
ugh
|dw:1433722130469:dw|
multiplication right?
Nope, perimeter would be the sum of all four sides.
Next case!
Please!
We only have two more cases.
I dont like this one ;-;
im not even sure where to start .-.
|dw:1433722487080:dw|
can you finish it?
\[\huge~\frac{ 2(x−2)(x^3+4x^2−12x+4) }{ (x+2)(x−2)(x−2) }\]
wow, not like that. The addition had already been done, because the denominators are the same.
All you need to do is to cancel (x-2) and give the answer!
|dw:1433722913558:dw|
>.<
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
Got it
good!
another?
no ;;
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.
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|
Once you can find the common denominator readily, rest is relatively straight-forward.
Makes sense :)
So that's good, we'll proceed with an example.
Simplify \(\Large \frac{x+1}{5} + \frac{2x}{6}\) Can you first try to find the common denominator?
|dw:1433726621112:dw|
Do you agree?
Yess
\[\huge~\frac{8x+3}{15} \]
Exactly, so we go with the same idea for the other rational expressions.
Now do you still remember equivalent fractions?
yes
for example, \(\Large \frac{9}{5x} = \frac{? }{40x^4 }\)
8?
It's basically cross multiplication, \(\Large ? =40x^4 ~\frac{9}{5x} = 72x^3\) Dose that make sense?
so \(\Large \frac{9}{5x}= \frac{72x^3}{40x^4}\)
is that ok? We need to use this step later on.
yes its fine
Ok, try to find the sum in its simplest form: \(\Large \frac{1}{x^2} + \frac{2}{x}\)
\[\huge~\frac{2x+1}{x^2}\]
Yes, that correct!
\[\large~\frac{ 3 }{ x }\]
cannot do that, because x and 1 are not like terms, so there is nothing to factor out.
oh right
Add and simplify if possible \(\Large \frac{1}{x+1} + \frac{1}{x-1}\)
\[\huge~\frac{2x}{\left(x+1\right)\left(x-1\right)} \]
Wow, are we wasting time?
Try this: add \(\Large \frac{x-5}{x+5} + \frac{x+2}{x-2}\)
Sorry, subtract!
\(\Large \frac{x-5}{x+5} - \frac{x+2}{x-2}\)
Use FOIL.
\[\huge~\frac{14x}{\left(x+5\right)\left(x-2\right)}\]
\(\Large \frac{x}{x-10} + \frac{x+4}{x+6}\)
wait i cant do this one .-.
When there is no common factor, you only have to "kind of" cross multiply.
\[\frac{2x^2-40}{\left(x+6\right)\left(x-10\right)} \] i have it uptil this part
|dw:1433728624180:dw|
Yes, it's correct, but need to factor out the 2 to give 2(x^2-20) at the numerator.
Another one?
nuu next case!
ok. Have you done rational functions?
p.678
11.7 Rational Functions ==============
Havent done that
Sure?
yes
ok, good. Together with the calculator(s), we have covered everything!
._. lol thanks for all the help :)
You're welcome! :)

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