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July 6th, 2017, 04:14 PM   #11
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Is this a good definition ?

Steps to Find the LCM of Two or More Rational Expressions

1. Factor all denominators completely.
2. The LCM is the product of unique prime factors from the denominators,
where each factor is raised to the highest power to which it appears in
any denominator.





Quote:
We have seen in Class VII that in algebraic expressions, terms are formed as products of
factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed
by the factors 5, x and y, i.e.,
5xy = 5 * x * y
Observe that the factors 5, x and y of 5xy cannot further
be expressed as a product of factors. We may say that 5,
x and y are ‘prime’ factors of 5xy. In algebraic expressions,
we use the word ‘irreducible’ in place of ‘prime’. We say that
5 × x × y is the irreducible form of 5xy. Note 5 × (xy) is not
an irreducible form of 5xy, since the factor xy can be further
expressed as a product of x and y, i.e., xy = x × y.

1. Factor all denominators completely.

4x = 2.2.x , x2 = x.x , 2x2 =2.x.x

So ,

2. The LCM is the "product" of unique ( being the only one of its kind; unlike anything else )"pairs of irreducible factors" from the denominators,
where each factor is raised to the highest power to which it appears in
any denominator.

4x = 2.2.x , x2 = x.x , 2x2 =2.x.x
So LCM = 2.2.x.x = 4x2
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July 6th, 2017, 06:20 PM   #12
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Quote:
Originally Posted by greg1313 View Post
Can you solve

$$\frac{x^2-4x+4}{x^2+4x-12}\ge0$$

for $x$?
$\frac{x^2-4x+4}{x^2+4x-12}$ factors as $\frac{(x-2)^2}{(x-2)(x+6)}$. The numerator's always
positive and the denominator is negative for $x\in(-6,2)$
and positive everywhere else. So the solution set is

$$x\in(-\infty,-6)\cup(2,\infty)$$

As there's a zero in the denominator at $x=2$, the original
function is never equal to zero - though it may appear that
way on a plot. Can you guess why?

If you have questions about the notation or anything else
I'd be happy to answer them. I hope this helped.
Thanks from awholenumber
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July 6th, 2017, 08:53 PM   #13
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Thanks a lot greg1313, I knew it involved factoring, but I didn't know the exact steps though.

Quote:
Originally Posted by greg1313 View Post
As there's a zero in the denominator at x=2, the original
function is never equal to zero - though it may appear that
way on a plot. Can you guess why?
No, I am not sure why.

As for the question,
Is this an OK definition?



Steps to Find the LCM of Two or More Rational Expressions

1. Factor all denominators completely.
2. The LCM is the product of unique prime factors from the denominators,
where each factor is raised to the highest power to which it appears in
any denominator.

4x = 2.2.x , x2 = x.x , 2x2 =2.x.x
So LCM = 2.2.x.x = 4x2

Quote:
The LCM is the "product" of the unique (being the only one of its kind; unlike anything else) "pairs of irreducible factors" from the denominators,
where each factor is raised to the highest power to which it appears in
any denominator.

Last edited by skipjack; July 10th, 2017 at 03:39 PM.
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July 6th, 2017, 11:17 PM   #14
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I like this definition better than the old one ,

To calculate an LCM for a rational function, follow these steps:
1. Factor all denominator polynomials completely.
2. Make a list that contains one copy of each factor, all multiplied together.
3. The power of each factor in that list should be the highest power that factor is
raised to in any denominator.
4. The list of factors and powers you generated is the LCM.



4x = 2.2.x , x2 = x.x , 2x2 =2.x.x
So LCM = 2.2.x.x = 4x2
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July 7th, 2017, 02:51 AM   #15
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You calculate the LCM by dividing the product of the two quantities by the greatest common factor (GCF) of the two numbers.

Example using 15 and 6:

$\displaystyle 15 = 3 \times 5$
$\displaystyle 6 = 2 \times 3$

3 is in both lists, therefore...

$\displaystyle GCF(15, 6) = 3$

$\displaystyle LCM = \frac{Product}{GCF} = \frac{15 \times 6}{3} = \frac{90}{3} = 30$

-----

To get the LCM of more than two quantities, you need to pick any two numbers, calculate the LCM using the above, then repeat the same process with the LCM result. Basically:

LCM(15, 6, 9) = LCM(30, 9) = 90

-----

You can do the above methods for variables too; we just assume that variables are like prime numbers and cannot be decomposed further.

Last edited by Benit13; July 7th, 2017 at 02:57 AM.
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July 7th, 2017, 08:15 AM   #16
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Thanks a lot .
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July 7th, 2017, 10:01 AM   #17
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Unfortunately I can't see any of the images you post, so I have no idea what your reply to my post#9 was, except that you replied.

Last edited by skipjack; July 7th, 2017 at 05:46 PM.
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July 7th, 2017, 10:06 AM   #18
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This PDF has a lot more explanations on how to find the LCM properly

Textbook | The Essential Elementary & Intermediate Algebra

http://www.jonblakely.com/wp-content/uploads/5_3.pdf
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July 7th, 2017, 10:09 AM   #19
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Yes but I wasn't discussing the LCM particularly.

I was addressing your opening post, some of which I don't fully understand.

What has x to do with electrons or binary logic etc?
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July 7th, 2017, 10:20 AM   #20
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Hello studiot, maybe postimage is blocked in your place.

Reposting it with imgur

I was just telling that this is the best picture of digital electronics I have come across ...





Here,

x is electrons
x is a quantity
x is binary
x is a point mass object
x is digital logic


Last edited by skipjack; July 7th, 2017 at 05:47 PM.
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