VCE Methods Question Thread!

[cosine]

Also, from the checkpoints 2014, a similar question is I did synthetic division and I STILL got it wrong, I mean, theres nothing to factor out, so whats going on?

[knightrider]

There is a method of this that will work in all cases - I will show you this method. However, there IS another method that will work brilliantly with this example that is a lot less effort - and I'll follow up the first method with this.

For the first method, let's put these simultaneous equations into matrix form:


We know that a solution for this will ONLY exist if the determinant of the first matrix is non-zero (or if x and y are 0, but that's trivial). So, if the determinant IS zero, then there is NO unique solution, and every other time, there IS a unique solution. So, let's find when the determinant is zero and discount those solutions:



So, there exists a unique solution when

On our second method, let's transpose both equations like so:



Now, we know that if the gradients are equal, then there are either infinitely many or no solutions. So, let's find out when the gradients are equal, and exclude those values:



Which reflects what we got earlier, so we once again say that


Note: the second method will always work as well, it's just that it's not as nice if the two lines don't intercept at (0, 0).

Thanks so much EulerFan101

was the second method the   method  that will work in all cases.
and the matrices method was the one that worked brilliantly with this example
also which book teaches you how to solve these using matrices(can you tell me page number as well as book if possible)

[keltingmeith]

Also, from the checkpoints 2014, a similar question is I did synthetic division and I STILL got it wrong, I mean, theres nothing to factor out, so whats going on?

Actually, there's a -1 you need to factor out. Remember, it HAS to be in the form (x-r), this is in the form of (r-x). (also, please use brackets from now on if you won't use the \frac command. Just makes our lives easier. ) Here's how you do your two example:

1)

First, factor the three out of the bottom:



Now, divide:

Code: [Select]

1/3 | 1
    |____
      1

1/3 | 1
    |____1/3_
      1  1/3
So, this means that 1/3 is a remainder, and we get:



2)

First, take out the negative 1:



Now, divide as per usual, and you should get -1 and 1, with 1 being the remainder. So, this means that:



Just for fun, I'm going to do this one with division by inspection, see if that helps you understand it at all:


We want the tops to match, so:


Continuing above:


And we see that our answers match.

Thanks so much EulerFan101

was the second method the   method  that will work in all cases.
and the matrices method was the one that worked brilliantly with this example
also which book teaches you how to solve these using matrices(can you tell me page number as well as book if possible)

Sorry, I worded this wrong. They'll both always work, it's just that the matrices method works better in all cases, and the second method was something you could easily do in this case.

[knightrider]

"Sorry, I worded this wrong. They'll both always work, it's just that the matrices method works better in all cases, and the second method was something you could easily do in this case."

Eulerfan  also which book teaches you how to solve these using matrices(can you tell me page number as well as book if possible)

[cosine]

2)

First, take out the negative 1:

So it always has to be in the form of (x-r), is this only the denominator (the factor)? And also, above, I see you took out -1 in order to make (2-x) into the form (x-r), but how/why did the numerator become -x+3?

Thanks so much, you never cease to amaze me!

[keltingmeith]

"Sorry, I worded this wrong. They'll both always work, it's just that the matrices method works better in all cases, and the second method was something you could easily do in this case."

Eulerfan  also which book teaches you how to solve these using matrices(can you tell me page number as well as book if possible)

I've got no clue, I don't own a textbook anymore. ~cue other helpful board member~

So it always has to be in the form of (x-r), is this only the denominator (the factor)? And also, above, I see you took out -1 in order to make (2-x) into the form (x-r), but how/why did the numerator become -x+3?

Thanks so much, you never cease to amaze me!

Yep - just the denominator needs to be in that form.

And the numerator changed because I moved the negative from the bottom to the top.

[cosine]

And the numerator changed because I moved the negative from the bottom to the top.
[/quote]
But why did you do that? Is it just because you factored it out, and it remains in front of the fraction, so you just multiplied the numerator by it? I dont know, it just seems confusing... lol

[IndefatigableLover]

"Sorry, I worded this wrong. They'll both always work, it's just that the matrices method works better in all cases, and the second method was something you could easily do in this case."

Eulerfan  also which book teaches you how to solve these using matrices(can you tell me page number as well as book if possible)

Page 58 in Essentials 3&4 for Matrices Method

[keltingmeith]

But why did you do that? Is it just because you factored it out, and it remains in front of the fraction, so you just multiplied the numerator by it? I dont know, it just seems confusing... lol

Because the negative HAD to go somewhere - if I left it on the bottom, I couldn't have divided. And if I just remove it, I change the question entirely. So, I moved it to the top.

[cosine]

Because the negative HAD to go somewhere - if I left it on the bottom, I couldn't have divided. And if I just remove it, I change the question entirely. So, I moved it to the top.

!

[cosine]

Can someone please help me with this:

Sketch:
 
So, you factor out the 2 in the denominator:


From now, I do synthetic division and I get -4, and 9 is my remainder, so the equation goes like this:


But for some reason, the book says the value of 'k' (-4) is actually -2? Can someone please....

[theshunpo]

Can someone please help me with this:

Sketch:
 
So, you factor out the 2 in the denominator:


From now, I do synthetic division and I get -4, and 9 is my remainder, so the equation goes like this:


But for some reason, the book says the value of 'k' (-4) is actually -2? Can someone please....

I can't help you with Synthetic Division (I've never used it before). I just did long polynomial division.

[brightsky]

I would steer clear of synthetic division entirely, as it is not a formal technique covered in Methods textbooks. By all means, use synthetic division to check your answer, but when showing working out, it is best to use either what EulerFan101 neatly called division by inspection, or polynomial long division.

Division by inspection:

As I've mentioned before, the aim of 'division by inspection' is to manipulate the numerator into the sum of a multiple of the denominator plus a remainder term.

(3-4x)/(2x+3) = [-(4x+6) + 9]/(2x+3) = [-2(2x+3) + 9]/(2x+3) = -2 + 9/(2x+3) = 9/(2x+3) - 2

As EulerFan101 mentioned, this technique is really useful, as it allows you to expand fractions of this sort in a matter of seconds. There is nothing particularly erroneous about the technique that you are using at the moment, it's just that I get the feeling you are making life difficult for yourself in insisting on using synthetic division.

Polynomial long division:

This technique, unlike synthetic division, is covered in most Methods textbooks and makes a lot more sense than synthetic division. Apologies in advance for the formatting.

             -2
            ---------
2x + 3 | -4x + 3
             -4x - 6
           ------------
                     9

Hence, (-4x + 3)/(2x+3) = -2 + 9/(2x+3) = 9/(2x+3) - 2.

Of course, you'll get the same answer if you apply synthetic division correctly, but you may not get full marks on the exam if you show synthetic division in preference to the two techniques listed above.

[cosine]

So youre telling me to drop synthetic division? I mean it works fine with just finding linear factors of polynomials, but for some odd reason it just wont work out when you need to use it for hyperbolas. How can I learn the Division by inspection and polynomial long division, because my text book (Heinemann) doesnt even show neither, not even synthetic division, but just assumes you already know it lol

Thanks for the help though, appreciate it

[DSubShell]

How can I learn the Division by inspection and polynomial long division,

I feel like both these skills are very "practical" based more than theory, so the best way to learn these skills would be through actually talking with a person: ie. Friend, Teacher or Tutor.

However, that might be hard in this period of the year, so what I suggest is search for "Long polynomial division" on google (or youtube) and see what comes up. I think a video tutorial would be the best as it would go through step by step.
Here is one from Khan Academy (usually makes good stuff):
https://www.khanacademy.org/math/algebra2/polynomial_and_rational/dividing_polynomials/v/polynomial-division
And here is a step-by-step text guide:
http://www.purplemath.com/modules/polydiv2.htm

But I think you should do the search and see what clicks with you. However, try your best to meet someone in real life and do it, so you get the skill down patt without any questions or flaws in your working.