VCE Methods Question Thread!

[Zahta]

thank you so much

[Zahta]

i still dont get the row echelon form.  how do i deduce it to know infinteltly many or no solution wow im so confused

[Insa]

I can't seem to factorise this expression. When I use long division, I always seem to get a remainder. Can anyone help me? Thanks.

[brightsky]

x=2 works.
so does x=-1.
it's all about trial and error. just plug in factors of 36 and cross your fingers.

[Insa]

Ahhh I see now. Thanks 

[Panicmode]

I can't seem to factorise this expression. When I use long division, I always seem to get a remainder. Can anyone help me? Thanks.

Have you heard of/used the remainder theorem?

If ax + b is a factor than it follows that

ax + b = 0

and by extension;

x = -b / a

Therefore you can substitute random values for x into original equation and find out whether they are factors without having to go through the entire process of long division.

For example, with the question you had previously;

2x^4 + 7x^3 - 31x^2 + 36

Try substituting in random numerals. Let's start with 1.

let f(x) = 2x^4 + 7x^3 - 31x^2 + 36

f(1) = 2 + 7 - 31 + 36
f(1) = 14

Therefore -b / a = 1

x - 1 is not a factor.

f(-1) = 2 - 7 - 31 + 36
f(-1) = 0

Therefore -b / a = -1

x + 1 is a factor.


From here you could divide 2x^4 + 7x^3 - 31x^2 + 36 by x + 1 or you could keep looking for factors.

[Bhootnike]

I can't seem to factorise this expression. When I use long division, I always seem to get a remainder. Can anyone help me? Thanks.

you probably forgot to add in the 0x . cause if you look at the degrees of x, it goes 4 3 2 0. so it should be:

[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... 

[Insa]

Thank you for your help guys. I got the question right this time 

[pi]

it's all about trial and error. just plug in factors of 36 and cross your fingers.

Nah, use some hardcore formulas (they actually work too!) for :





But seriously, brightsky's method is better

[TrueTears]

http://en.wikipedia.org/wiki/Rational_root_theorem suffices for vce methods most of the time.

[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

[Zahta]

can someone please help me with no soltuion and infinitely many solutions with matrices/ simultanous equations. I am really confused with them  like with the equation mx+2y=8 and 4x-(2-m)y=2m

[pi]

can someone please help me with no soltuion and infinitely many solutions with matrices/ simultanous equations. I am really confused with them  like with the equation mx+2y=8 and 4x-(2-m)y=2m

If it confuses you, don't do it via matrices It confused the hell out of me, so I always did it via simultaneous equations:



For there to be infinite solutions, and

For no solutions, and


Work from there

(I hope I haven't forgotten everything and stuffed up though )


edit: made obvious minus mistakes...

[Panicmode]

Okay, although you don't have to use matrices, it is by far the easiest method once you understand it.

The theory behind this comes from setting out the two equations in matrix form to solve. In this case, you write both equations, one on top of the other and make sure the test variables (x and y) align with the respective counterpart in the other equation. Then you go from left to write and fill in the blanks in your matrix.

Once the matrix has been constructed, the next step would of course be to rearrange the equation so that the X matrix is by itself. This involves finding the inverse matrix of A.

To find the inverse of a 2x2 matrix, the top left and bottom right switch places, the top right and bottom left switch signs, and the whole thing is multiplied by 1 on the determinate of A.

The determinant of A [det(A)] is calculated by multiplying the top right with the bottom left and subtracting it from the bottom right multiplied by the top left.

It goes to stand that if the det(A) = 0, then the inverse of A doesn't exist.


So, all that really need be done in the first place is find det(A), set it equal to zero and solve.

See attached scan.