VCE Methods Question Thread!

[Zahta]

i finally underastand it thankyou so much , i really appreciate it . thanks alot

[Bhootnike]

quick question,
is synthetic division appropriate enough to use for 'showing' or 'proving' things which involve long division ? as in, show bla bla bla is divisible for any value of x... 

I don't think its an accepted VCAA method, having said that its fine to use it to check your answer. I've also not seen a "show via long division" question before, so it shouldn't matter anyway

there was a question in essentials :
Prove that the expression   is divisible by( x + 1) for all values of k.
using synethetic division, its rather easy. takes literally half a minute.

                  | 1   k-1  k-9    -7
             -1  |     -1   -k+2    7
                  |____________
                    1    k-2    -7    0

if there was a similar question on an exam, could you just do that, ? and you know, add in the, "hence, all values for ..... bla bla etc etc"
i found using long div on this longer, and i screwed it up. :S


Moderator action: removed real name, sorry for the inconvenience

[pi]

quick question,
is synthetic division appropriate enough to use for 'showing' or 'proving' things which involve long division ? as in, show bla bla bla is divisible for any value of x... 

I don't think its an accepted VCAA method, having said that its fine to use it to check your answer. I've also not seen a "show via long division" question before, so it shouldn't matter anyway

there was a question in essentials :
Prove that the expression   is divisible by( x + 1) for all values of k.
using synethetic division, its rather easy. takes literally half a minute.

                  | 1   k-1  k-9    -7
             -1  |     -1   -k+2    7
                  |____________
                    1    k-2    -7    0

if there was a similar question on an exam, could you just do that, ? and you know, add in the, "hence, all values for ..... bla bla etc etc"
i found using long div on this longer, and i screwed it up. :S

Yeah, it may be faster, but its not accepted by VCAA Not even all teachers/assessors know what synthetic division is tbh.

You'd need to do long division.

Moderator action: removed real name, sorry for the inconvenience

[Zahta]

Also thank you pi .

Moderator action: removed real name, sorry for the inconvenience

[Hutchoo]

quick question,
is synthetic division appropriate enough to use for 'showing' or 'proving' things which involve long division ? as in, show bla bla bla is divisible for any value of x... 

I don't think its an accepted VCAA method, having said that its fine to use it to check your answer. I've also not seen a "show via long division" question before, so it shouldn't matter anyway

there was a question in essentials :
Prove that the expression   is divisible by( x + 1) for all values of k.
using synethetic division, its rather easy. takes literally half a minute.

                  | 1   k-1  k-9    -7
             -1  |     -1   -k+2    7
                  |____________
                    1    k-2    -7    0

if there was a similar question on an exam, could you just do that, ? and you know, add in the, "hence, all values for ..... bla bla etc etc"
i found using long div on this longer, and i screwed it up. :S

Yeah, it may be faster, but its not accepted by VCAA Not even all teachers/assessors know what synthetic division is tbh.

You'd need to do long division.

Is synthetic division srsly not accepted? NOOOOOOOOOOOOOO.

Moderator action: removed real name, sorry for the inconvenience

[Bhootnike]

we're doomed.

[pi]

Nope, not accepted

[Bhootnike]

well. it still enables you to rapidly do long div! so you could do it to know your answer, then just scribble down some of those numbers in long div. hopefully examiners wont check, will see the right answer, tick !    boss.

[pi]

Yeah, no. Assessors aren't stupid and in a "prove" question like the one you gave, there isn't much to check anyway as you already know the "answer"...

Just do long div, after practise it does't take that long It seriously won't!

[Planck's constant]

Nope, not accepted

Yeah, no. Assessors aren't stupid and in a "prove" question like the one you gave, there isn't much to check anyway as you already know the "answer"...

Just do long div, after practise it does't take that long It seriously won't!

Personally, it would never occur to me to do any sort of division on this problem. Long, short, medium or synthetic.
They give you a BIG HINT !
(x+1) is a factor
Use it
Factorise

Moderator action: removed real name, sorry for the inconvenience

[kamil9876]

^ good point. Even easier than factorization: you know that if is a polynomial, then divides it if and only if

[rife168]

would someone be able to go through indefinite integrals where the result is
I understand it intuitively and can get most of the questions, but I'm not 100% sure of how to use the absolute values and whatnot.

[Panicmode]

would someone be able to go through indefinite integrals where the result is
I understand it intuitively and can get most of the questions, but I'm not 100% sure of how to use the absolute values and whatnot.

take this an example.




thats all.

Your example is wrong. For antidifferentiation to be possible, the derivative of the bottom must be found on top. This is achieved by multiplying by a factor of The "a" is multiplied through and then a factor of is taken out.

Therefore the correct answer is

[rife168]

would someone be able to go through indefinite integrals where the result is
I understand it intuitively and can get most of the questions, but I'm not 100% sure of how to use the absolute values and whatnot.

take this an example.




thats all.

Your example is wrong. For antidifferentiation to be possible, the derivative of the bottom must be found on top. This is achieved by multiplying by a factor of The "a" is multiplied through and then a factor of is taken out.

Therefore the correct answer is

Yeah, that makes sense, it's the use of the absolute value that is confusing me a little.
The explanation in Essentials isn't very clear, and the arithmetic in one section seems incorrect, it states  

[Panicmode]

would someone be able to go through indefinite integrals where the result is
I understand it intuitively and can get most of the questions, but I'm not 100% sure of how to use the absolute values and whatnot.

take this an example.




thats all.

Your example is wrong. For antidifferentiation to be possible, the derivative of the bottom must be found on top. This is achieved by multiplying by a factor of The "a" is multiplied through and then a factor of is taken out.

Therefore the correct answer is

Yeah, that makes sense, it's the use of the absolute value that is confusing me a little.
The explanation in Essentials isn't very clear, and the arithmetic in one section seems incorrect, it states  

The reason absolute values is used is to account for the possibility of negative values of x. Since the original function has not got a restricted domain, both negative and positive values can be used. The only value that cannot be used would be one that causes the denominator to equal 0. When antidifferentiating, one must therefore account for this - by using the absolute value signs as it is not possible to have a rational log of a negative number.

Understand?