Re: simplify

On 15 Jun 2007 13:35:18 +1000, Gary Wessle <phddas@xxxxxxxxx> wrote:

quasi <quasi@xxxxxxxx> writes:

On 15 Jun 2007 05:40:02 +1000, Gary Wessle <phddas@xxxxxxxxx> wrote:

quasi <quasi@xxxxxxxx> writes:

On 14 Jun 2007 16:28:28 +1000, Gary Wessle <phddas@xxxxxxxxx> wrote:

quasi <quasi@xxxxxxxx> writes:

On 14 Jun 2007 07:05:39 +1000, Gary Wessle <phddas@xxxxxxxxx> wrote:

quasi <quasi@xxxxxxxx> writes:

On 14 Jun 2007 01:54:42 +1000, Gary Wessle <phddas@xxxxxxxxx> wrote:

quasi <quasi@xxxxxxxx> writes:

On 13 Jun 2007 19:32:06 +1000, Gary Wessle <phddas@xxxxxxxxx> wrote:

Hi
I need some guide on what to think about in-order to simplify this

(1/( (a-b)(a-c) )) + (1/( (b-c)(b-a) )) + (1/( (c-a)(c-b) ))

if I were to multiply all the denominator to go the lowest common
denominator I get, (-a^2-b^2)(-b^2-c^2)(-a^2-c^2)

I feel I am making it more complicated than I have to.

If you find the correct LCD, it will get very simple very quickly.

Hint: Find the LCD, but leave it in factored form.

quasi

here is another one, should I start another thread or stay here.

I can simplify each of brackets or multiply them out, or there is
another trick?

( 1 + 45/(x-8) - 26/(x-6) ) ( 3 - 65/(x+7) + 8/(x-2) )

Definitely simplify each of the 2 factors first. Leave the common
denominators in factored form in the hope of potential cancellation
with a factor of one of the new numerators. Thus, after combining each
factor over an LCD, make sure to try to factor the new numerators.

the LCD for the first term will be (x-8)(x-6) and for the second term
will be (x+7)(x-2). am I correct?

Yes, for the first and second _factors_.

( ((x-8)(x-6) + 2) / (x-8)(x-6) ) the second term does not look as
friendly. am I on?

I have no idea what the above represents, but the recommendation was
to simplify (combine and refactor) each of the original factors
separately.

Thus, you want to combine the factors:

( 1 + 45/(x-8) - 26/(x-6) )

and

( 3 - 65/(x+7) + 8/(x-2) )

Each of the above factors has its own LCD. Combine the fractions using
the LCD, but leave the LCDs in factored form. Also, simplify (expand)
the new numerators and then try to factor them.

Once each of the 2 above factors is fully simplified (combined and
refactored), the next step will be obvious.

ok, here we go.
the first factor, (expand & refactor / LCD) will be
( (x+7)(x-2) ) / ( (x-8)(x-6) )
good, now this numerators looks identical to the denominator of the
second factor, now the second factor I have some trouble with.

the second factor (expand / LCD) will be
(3x^2 - 32x + 144) / ( (x+7)(x-2) )

Check your work.

3x^2 - 32x + 144 is not correct.

your are right
the correct value is
3x^2 - 42x + 144
but that still leaves me wondering how to go from here, I mean how to
factor this value above to continue with my original problem.

Look at the coefficient of x^2.

The factors, if any, would need to have what leading terms? That
should get you started.

Also, use a little reverse psychology.

What factors would you like to have, if you could get so lucky?

man, I've got a bad imagination!
it is 3(x-8)(x-6)
besides,
say I am not so lucky, and the reverse psychology is not so
successful, is there a systematic way to do those kind of factorization?

Good question.

There are several ways.

For one thing, As [Mr.] Lynn Kurtz points out, we both missed the
common factor of 3. A common factor is the first thing to look for --
if you find one, remove it, and try to factor what remains.

For quadratic polynomials, there are a number of factoring methods.

Suppose the quadratic is ax^2+bx+c, where a,b,c are integers.

Assume gcd(a,b,c)=1, that is, a,b,c have no common factor (since if
there was a common factor, you would have removed it in advance).

Your goal is to write

ax^2+bx+c = (a1*x+c1)(a2*x+c2)

Clearly, we must have a1*a2=a and c1*c2=c.

If a,c are easily factored and if they have just a small number of
factor pairs, there's usually no point to getting fancy -- just use
trial pairs of factors for a and c and test to see which combination
works.

When the quadratic is monic (a=1) the situation is simpler. In that
case, You want to write

x^2+bx+c=(x+c1)(x+c2)

This implies c1*c2=c and c1+c2=b.

Thus, for c1,c2, you want 2 numbers whose product is c and whose sum
is b. Again, if c has a small number of pairs of factors, trial and
error is usually the simplest way.

When factoring a and c is too hard or gives so many possibilities that
trial and error might be too tedious, there are a number of other
methods.

One method which always works is to complete the square. As an
example, use the problem you had ...

Factor 3x^2 - 42x + 144

First look for a common factor of all the coefficients. The only
possible common factor (greater than 1) is 3. Trying it, we see that
42 and 144 are both multiples of 3, which gives

3*(x^2-14x+48)

Actually, for the completing the square method, we always factor out
the leading coefficient, even if you end up with fractional
coefficients for the remaining quadratic.

In any case, you want to factor

x^2-14x+48

Of course, you might see right away that

x^2-14x+48=(x-6)(x-8)

But suppose you don't see it, and suppose you don't feel like trying
all possible pairs of factors of 48.

Let's try the completing the square approach. Thus,

x^2-14x+48
=(x^2-14x+49) - 49 +48
=(x^2-14x+49) - 1
=(x-7)^2 - 1
=(x-7+1)(x-7-1)
=(x-6)(x-8)

Then bring back the factor of 3 that was "on hold", so you get

3(x-6)(x-8)

Note -- the constant 49 was obtained (mentally) by the steps:

14/2=7
7^2=49

That's the standard trick for completing the square of a monic
(leading coefficient = 1) quadratic.

As I mentioned, there are other methods, but the method of completing
the square is used for lots of different types of math problems, so
applying it here will help you get familiar with it.

quasi
.

Relevant Pages

  • Re: simplify
    ... I need some guide on what to think about in-order to simplify this ... if I were to multiply all the denominator to go the lowest common ... If you find the correct LCD, it will get very simple very quickly. ... there are a number of factoring methods. ...
    (sci.math)
  • Re: LCD Mounting
    ... commonly available, pre-fab plastic housings? ... so it's not like they're thinking the enclosure is going to be ... Preferably from one of the better known LCD manufacturers. ... I've now tried about 15 common LCD and NONE of them ...
    (sci.electronics.design)
  • Re: No constitutional rights in Oregon?
    ... The problem isn't so much with the legal precision a well-crafted law ... "common sense," and with it, a life that might be lived more freely. ... Thoreau was right: "Simplify, simplify!" ... Perhaps regrettably, Thoreau ...
    (rec.pyrotechnics)
  • Re: LCD drivers
    ... >>>It has more than enough segment drivers. ... The LCD has not yet been sourced. ... version of the micro you can use try it out on some LCDs. ... development version and compare against 8 common variant and 4 common ...
    (comp.arch.embedded)
  • Re: Finding triples (a,b,c) such that a^2 + b^2 + c^2 = P^2 and abc = Q^3
    ... power of p dividing is not p^for some integer k. ... Let C be the smallest integer so that abC is a third power, ... ignore values c where k has a common divisor with both a and b. ... The algorithm could be made a lot faster by factoring each b, ...
    (sci.math)
Loading