VCE Methods Question Thread!

[keltingmeith]

Thanks Brightsky
Thanks DSubShell  could you please explain the concept of using inspection

We'll use your earlier example, f(x)=x^3-3x-2.

Now, we know that f(2)=0, so f(x)=(x-2)q(x), the question we want to answer is what q(x) is. We know that q(x) should be some polynomial, and that it should be one degree lower than f(x), so we get:

f(x)=(x-2)(ax^2+bx+c)

Expanding this out, we get:

f(x)=ax^3+bx^2+cx-2ax^2+2bx-2c

Now, by equating coefficients, we can immediately tell that a=c=1, so we get:

f(x)=(x-2)(x^2+bx+1)

Finally, we need to solve for b - by equating coefficients again, we can see that:

2b+c=-3
2b=-3-1
b=-2

So, f(x)=(x-2)(x^2-2x+1)=(x-2)(x-1)^2

Doing it "by inspection" is that method, however instead of expanding, equating coefficients, etc. you do it all in your head, focusing on the constant and the leading coefficient to ascertain two of the unknowns, and then figuring out the last few by logic.

Remember that polynomial division is also a suitable technique for reducing a polynomial into its factors.

[knightrider]

We'll use your earlier example, f(x)=x^3-3x-2.

Now, we know that f(2)=0, so f(x)=(x-2)q(x), the question we want to answer is what q(x) is. We know that q(x) should be some polynomial, and that it should be one degree lower than f(x), so we get:

f(x)=(x-2)(ax^2+bx+c)

Expanding this out, we get:

f(x)=ax^3+bx^2+cx-2ax^2+2bx-2c

Now, by equating coefficients, we can immediately tell that a=c=1, so we get:

f(x)=(x-2)(x^2+bx+1)

Finally, we need to solve for b - by equating coefficients again, we can see that:

2b+c=-3
2b=-3-1
b=-2

So, f(x)=(x-2)(x^2-2x+1)=(x-2)(x-1)^2

Doing it "by inspection" is that method, however instead of expanding, equating coefficients, etc. you do it all in your head, focusing on the constant and the leading coefficient to ascertain two of the unknowns, and then figuring out the last few by logic.

Remember that polynomial division is also a suitable technique for reducing a polynomial into its factors.

Thanks Eulerfan101 which methods do you think is the most effective

[keltingmeith]

Thanks Eulerfan101 which methods do you think is the most effective

The one you're fastest at and make the least amount of mistakes doing.

[lzxnl]

Also, if you ask me, a stupid way of finding inverses next to the adjoint method.

I'd do row operations over using the adjoint method on almost any day of the week

The one you're fastest at and make the least amount of mistakes doing.

Doing it by inspection is always going to be faster, but it'll take practice.

Again, I'll use the given example to show my own thought process when doing it 'by inspection' (in my head of course )

So x^3 - 3x - 2 = (x-2)(x^2 + ax + b)
Looking at the constant term, we see b = 1. Now, as both sides are identical for all values of x, they must be equal for, say x = 1. So I'm going to plug this into both sides. The right hand side is -4 and the left hand side is -1(1+a+1) = -(2+a) = -4. => a=2
x^3 - 3x - 2 = (x-2)(x^2 + 2x + 1) = (x-2)(x+1)^2

We'll use your earlier example, f(x)=x^3-3x-2.

Now, we know that f(2)=0, so f(x)=(x-2)q(x), the question we want to answer is what q(x) is. We know that q(x) should be some polynomial, and that it should be one degree lower than f(x), so we get:

f(x)=(x-2)(ax^2+bx+c)

Expanding this out, we get:

f(x)=ax^3+bx^2+cx-2ax^2+2bx-2c

Now, by equating coefficients, we can immediately tell that a=c=1, so we get:

f(x)=(x-2)(x^2+bx+1)

Finally, we need to solve for b - by equating coefficients again, we can see that:

2b+c=-3 you mean -2b + c? The first bracket is (x-2) not (x+2)
2b=-3-1
b=-2

So, f(x)=(x-2)(x^2-2x+1)=(x-2)(x-1)^2

Doing it "by inspection" is that method, however instead of expanding, equating coefficients, etc. you do it all in your head, focusing on the constant and the leading coefficient to ascertain two of the unknowns, and then figuring out the last few by logic.

Remember that polynomial division is also a suitable technique for reducing a polynomial into its factors.

[keltingmeith]

I'd do row operations over using the adjoint method on almost any day of the week

Weirdo.

And whoops - mah bad, should've paid more attention to what I was doing, hahah.

[brightsky]

I'd do row operations over using the adjoint method on almost any day of the week

So which days would you prefer the adjoint method instead?

[flares]

Hi everyone,
Does anyone here have Essential Maths Methods Units 3&4 worked solutions book and is wiling to sell it?

Thanks
also, i need a new methods tutor...

[lzxnl]

So which days would you prefer the adjoint method instead?

Those in which I have a 2x2 matrix and forget the basic inverse formula

Weirdo.

And whoops - mah bad, should've paid more attention to what I was doing, hahah.

The problem is, the adjoint method requires you to work out n (n-1)*(n-1) determinants and those take too much effort. And then you need the large nxn determinant too. So much effort.

Hi everyone,
Does anyone here have Essential Maths Methods Units 3&4 worked solutions book and is wiling to sell it?

Thanks
also, i need a new methods tutor...

If you want a new methods tutor, I suggest you post in the tutoring section. At least, make it clear where you're from as geographical location is very important. I wouldn't ever tutor a student that lived two hours away from me because simply put, I don't have that much time.

[flares]

If you want a new methods tutor, I suggest you post in the tutoring section. At least, make it clear where you're from as geographical location is very important. I wouldn't ever tutor a student that lived two hours away from me because simply put, I don't have that much time.

Thanks for the tip, buddy.
And also $$$ for petrol if it was two hours.

[keltingmeith]

And also $$$ for petrol if it was two hours.

Ignoring that, it's also a massive inconvenience to people who either:

a) Are full time and have other students they need to tutor,
b) Are part time because they're also studying/working another job.

I can't think of anybody who would travel two hours to tutor one kid, unless they were getting some astronomical amount per hour.

Those in which I have a 2x2 matrix and forget the basic inverse formula

Awww, you hurt me so.
Nah, for 3x3 and 4x4 I reckon the calculations aren't too bad, and find I can figure out those faster (particularly with a couple of zero entries. ) than I can reduce two matrices.

[flares]

Ignoring that, it's also a massive inconvenience to people who either:

a) Are full time and have other students they need to tutor,
b) Are part time because they're also studying/working another job.

I can't think of anybody who would travel two hours to tutor one kid, unless they were getting some astronomical amount per hour.

Good point. Okay, i know  lzxnl said to check out the tutoring section, but does anyone have a current methods tutor that they would recommend?
It's just sometimes the ones people recommend are good..

[pi]

As lzxnl rightly said, please keep this thread to maths. If you want a tutor, post in here Wanted

[lzxnl]

I can't think of anybody who would travel two hours to tutor one kid, unless they were getting some astronomical amount per hour.

Awww, you hurt me so.
Nah, for 3x3 and 4x4 I reckon the calculations aren't too bad, and find I can figure out those faster (particularly with a couple of zero entries. ) than I can reduce two matrices.

Kid you not, there was a student who emailed me asking for tutoring at his place which is in Altona. I live about 15 km east of the city. Odds that I'd take him on?

You'd need a lot of zeros for row expansion to be faster for a 4x4

[AirLandBus]

The one you're fastest at and make the least amount of mistakes doing.

So, in a exam situation, would they specifically state to use the inspection method? What are the chances of it occurring? Or would they always give you the option in which you want to solve the problem?

[keltingmeith]

So, in a exam situation, would they specifically state to use the inspection method? What are the chances of it occurring? Or would they always give you the option in which you want to solve the problem?

To memory, neither is specifically stated in the study design. However, they have so many tougher things they can ask, as opposed to "factorise this". No, the only reason you learn about this stuff is to graph the functions by hand, I doubt VCAA will ever specifically ask about factorising, rather going straight for the "graph f(x)".