Math Is Fun Forum

Hi SlowlyFading!

Hey, you are using LaTex!  That's fantastic.  Takes a little bit of getting used to but it's worth it.

17)  What are (5-6) and  (-2+2)?  And notice that the x term is  just -3x.  Nothing combines with it.
       Looking at the LaTex for your answer to 17) I can't figure out why the 2x^2 and 2x got stuck
together.
       

18)  Check!,  OK,  Correct,  Right,  Correctomundo!

19)  Looking at the LaTex I can't figure out why the extra x  got stuck to the 4x^2 term.
       Anyway without that extra x (should be a space where the x is) the answer looks good:
       

Notice that the \; causes the spaces in the LaTex output.  The more you put, the wider the space.

Help bobbym!  Can you help us figure out what the problem is with the LaTex? 

Hi SlowlyFading!   14 and 16 were correct.  The others just slightly off.

I have trouble following the input to LaTex (You can see it if you click on the "pretty math" output).
It's much easier to see what's going on in the pretty output.  Maybe that is part of the problem if
you are looking at the ^ notation as you are trying to work the problem.  Writing it by hand
like the pretty LaTex output might be easier to work with.

It's really easy to do the LaTex input for these.  You are ALREADY typing the correct input lines.
You just need to put (math) before and (/math) after BUT use [...] instead of (...).  All the \\ does
is create a new line.  LaTex does the correct spacing in its output automatically, so the spaces that
you insert into the input don't really count.  Use spaces there to make it look better to you.
Compare your line to the first line below it by looking at your line and then left clicking on the next
line.  You will see that they are essentially the same except for the = and \\'s causing new lines.

To do this LaTex you need to do the "Post a reply" not the "Quick post."  That way you have the
opportunity to click the "preview" button to see how it is going.

17. (5x^4 – 2x^2) – (3x – 2x^2 - 4x^3 + 6x^4)

18. (-x^2 + 4x - 3) + (x2 – 2x + 6)

19. (-4x^3 – x2 + 8) + (5x^2 - x – 12)

You'll notice that in going from the first "pretty" line to the second "pretty" line that I started with
the highest power available and scanned both polynomials for the terms with that power and then
put both of them next to each other  (if there were some in both polynomials).

HANG IN THERE!!!  You are well on the way. 

Good mornin' bobbym!

It's almost like starting a new thread I suppose.  Math is Fun has enough threads to weave a sweater!

Have a fantastic day!

Hi again Xerxes!

From your post #6 it looks like you are trying to do something like
    12        12    12     12
______ =  __ + __ + __  which is not legitimate.  Left side = 12/9  and right side = 4+3+6=13.
 
3+4+2      3      4       2

In the division algorithm the only term that we estimate (divide by) with is the first term.   In your
division case it is the x squared.   The x and the 2 are never used in estimations.  It might help if
you review some of the arithmetic division problems like  5785/523  in which only the 5 is the
estimator at each step.  The 2 and 3 are used in the multiplication steps but not in the estimations.

Perhaps you are confusing the above fraction with
3+4+2      3     4        2
______  = __ + __ +  __  which is OK to do.   It doesn't work with the fraction "flipped over."

    12        12    12      12

Hang in there!  You'll get it.

Hi JaneFairfax!

That's a very interesting approach.  I'll remember that.

And Xerxes,  polynomial division works pretty much like division in arithmetic.  Let's see if I can get
the numbers to line up fairly nicely in a post.  I'll just use the coefficients.

The 1 of the divisor 1  1  2 is the estimator for the process.

                      3  2  -1             Estimate 1 into 3 to get the 3
                   _____________
       1  1  2  |   3  5  7  3  -2     
                      3  3  6              multiply the 3 times 1  1  2 to get this line
                      ______
                           2  1  3       Subtract the 3  3  6 from the 3  5  7 to get 2  1 and bring down the 3
                           2  2  4       Estimate 1 into 2 and multiply 1  1  2 by 2 to get 2  2  4
                           ______         
                              -1  -1  -2    Subtract 2  2  4 from 2  1  3 to get - 1  -1 and bring down the -2
                              -1  -1  -2    Estimate 1 into -1 and multiply 1  1  2 by -1 to get -1  -1  -2
                              ________
                                         0    Subtract to get the remainder of zero.

If the remainder line had been something like 2  -3  then this would be 2x-3 in long form.

The basic algorithm is "estimate (division), then multiply, the subtract just like in arithmetic,
except now we are dealing with signed numbers.  If you get double digit numbers in the process
you cannot carry between columns line in arithmetic.

This is a "repeated subtraction" algorithm and can be done in a short (place value) form since
the polynomials are place value in an unknown base x.  But the process is basically the same
as that done in arithmetic except for now dealing with signed numbers.

And by the way, if you do get an estimation like 3 into 2 in a polynomial division then the number
to write up top is the fraction 2/3.  This makes the following arithmetic quite messy most of the
time, but we have no choice since we can't "borrow" in unknown base arithmetic.

Most authors of algebra books avoid these nasty problems like the plague.  They are quite happy
for the most part to "rig" the problems to be nice.  It makes no difference to a program like
Maple whether the numbers "turn out nice" or not.  It just crunches it out.  But to us humans it
makes a big difference.

Hi everyone!

The "-" can go either way.  If it is between two numbers/letters like 2x-3 then this is a subtraction
and the terms involved in the subtraction are 2x and 3.  Hence the "-" does not go with the 3.

On the other hand if we apply the definition of subtraction to get 2x+(-3) then the two terms
being added are 2x and -3.  Hence the "-" goes with the three. 

If all the subtractions are written as additions of opposites, then the terms (including their signs)
can be moved about indiscriminantly since addition is both commutative and associative.
Hence, as Bob said you can rearrange the terms if you "keep the sign with the term."

   3x  -  y + 7z  - 6a  -  2b
= 3x+(-y)+7z+(-6a)+(-2b) by applying the definition of subtraction
= 3x+7z+(-y)+(-6a)+(-2b) putting "like" terms together. 
= 3x + 7z  -y  -  6a  -  2b

Thanks, Bob

I'll check into it and probably some free ones too.

Hi y'all

To just find out whether a number n is prime or not, one of the fastest methods is as follows:

   In the following b is any base: 2,3,4,5,... preferably a prime.  I usually use 2 and do the
calculations on Maple.

   Calculate  b^(n-1) mod n.  If this turns out to be 1, then the chances are overwhelming
that it is prime.  The exceptions are called Carmichael numbers and they are they are few
and far between.  The first two are  651 and 1105. 
   41041 is the eleventh Carmichael number.  The factors of a psuedoprime have to be "just right"
for them to produce a 1 in this calculation. 

   Google Carmichael Number to see several articles about this.  Some of these articles should tell
how to determine whether of not you have a Carmichael number.

If n is a Carmichael number and gcd(n,b)>1 then the mod will not be 1.
There are only 8241 Carmichael numbers less than 10^12.  That's 8.241*10^(-7) %.

So this mod test is a quick test to determine primality with a high degree of probability. 
You might also Google the Miller-Rabin test for primality.  (There's a Wiki article.)

Primality is much easier and faster to determine than doing a factorization. 

Thanks phrontister!
That copy and paste worked OK.  I'm not sure what I was doing that didn't work.  I think I was using
the quick post and then copying from it and trying to go to the post reply.  At any rate I was jumping
around amonst pages and lost something in the process. 

Yeah, I like the x symbol for the multiplication better than the *.  Thanks for the suggestion. 

Hi SlowlyFading!
There's no reason to be slowly fading here.  There is an easier way to write and work these.

are place value like in arithmetic.  They can be rewritten as 8  27  -9  and 6  20  1 leaving spaces
between the coefficients since they go with different powers of x.  Line them up just like in
arithmetic and add or subtract.

8    27   -9
6    20    1
_________  and get the sum and difference like in arithmetic except you have signed numbers.

14  47   -8   is the sum.            ...NEVER... carry or borrow between columns because the base
                                              is unknown so we can't tell how much we are borrowing, etc.
2    7   -10  is the difference     This makes the algebra easier than the arithmetic.

Put the powers of x back in like you would powers of 10 (squared, first power, constant)

are the sum and difference, respectively.

If there are missing powers of x in the polynomial then you must put zeros for the coefficients
in the short form.

This is somewhat like asking someone to write you a check for $31 and seeing if they mind you
slipping in a couple of zeros ($3001).  The zeros make a difference.

We are taught quite well to do short forms in arithmetic, but in algebra they go back to the
expanded forms.  That makes in all harder: Try to  multiply 497*859 in the form

writing out all the exponents of 10 throughout.  How much fun would that be?

I hope this will help you and not confuse you.  Consider it like the options I saw for dinner in
a restaurant:  1) Take it.  2) Leave it. 

Hi phrontister!

Thanks a bunch for showing me the link and you too bobbym for showing phrrontister.  Seems to be
just what I was looking for!    Yeah! 

Hi bob!
Thanks for the input.
True.  One might have to do some 'splainin' to make sure the difference between regular multiplication
of fractions and enlarging are understood.  It could save a bit of writing when the argument is large:

2x+3 *      2
------    (3x  - 4x + 7)  instead of
3x+4 *         
               2
(2x+3)(3x - 4x + 7)
---------------------- 
               2
(3x+4)(3x - 4x + 7)

The LaTex for the complex fraction 3/3 over 6/3 has 31 characters if I have counted right.  The
output of the Latex is only 7 characters.  So the two dimensional output is much easier to read
and understand than the one dimensional Latex input.

Do you know of a good "front end" for Latex that could be used easily to get the LaTex code for this
forum?  I know there are several programs that allow inputting the math in a mathematical format
but I don't know of one that then allows you to view the equivalent LaTex.