Someone help me with this math problem...

[GreecePwns]

I need to know this by tommorrow (my teacher was absolutely no help at all, which means no one in my class knows). Here's an example:

Factor completely

(9x^[2y+2]) - (27x^2y)

Winner gets...i don't know...a cookie or something. Maybe a 1v1 when I get a free space.

Damn thats easy.

9x^2y(x^2-3x)

Remember, when you times polynomials you times the base, and add teh powers.

9x^2y times x^2 = 9x^(2y+2)

9x^2y times -3x = -(27x^2y) [Remember, -3x is actually -3x^1]

You're subtracting them, not multiplying.

No I'm not. When expanding the bracket, you do not subtract first. You times the first product by the outside product, and then you times the second product (with the negative sign) by the outside product.

Its impossible to simplify a subtraction which involves polynomials of different powers.

Ohhh...sorta got confused, but I knew it all along.

Thanks people.

[Titanic]

9X Log (2y + 2) = 27X Log (2y)

1/3 log(2y + 2) = Log (2y)

...

Howd you get an equals sign?

Btw, some advice, even if the first bit was right, the accepted answer would be "log(2y + 2) = 3Log (2y)". Always try to keep whole numbers, and avoid fractions, there very nasty.

[Titanic]

I need to know this by tommorrow (my teacher was absolutely no help at all, which means no one in my class knows). Here's an example:

Factor completely

(9x^[2y+2]) - (27x^2y)

Winner gets...i don't know...a cookie or something. Maybe a 1v1 when I get a free space.

Damn thats easy.

9x^2y(x^2-3x)

Remember, when you times polynomials you times the base, and add teh powers.

9x^2y times x^2 = 9x^(2y+2)

9x^2y times -3x = -(27x^2y) [Remember, -3x is actually -3x^1]

You're subtracting them, not multiplying.

No I'm not. When expanding the bracket, you do not subtract first. You times the first product by the outside product, and then you times the second product (with the negative sign) by the outside product.

Its impossible to simplify a subtraction which involves polynomials of different powers.

Ohhh...sorta got confused, but I knew it all along.

Thanks people.

Np....10th grade...how old are you?

[GreecePwns]

I need to know this by tommorrow (my teacher was absolutely no help at all, which means no one in my class knows). Here's an example:

Factor completely

(9x^[2y+2]) - (27x^2y)

Winner gets...i don't know...a cookie or something. Maybe a 1v1 when I get a free space.

Damn thats easy.

9x^2y(x^2-3x)

Remember, when you times polynomials you times the base, and add teh powers.

9x^2y times x^2 = 9x^(2y+2)

9x^2y times -3x = -(27x^2y) [Remember, -3x is actually -3x^1]

You're subtracting them, not multiplying.

No I'm not. When expanding the bracket, you do not subtract first. You times the first product by the outside product, and then you times the second product (with the negative sign) by the outside product.

Its impossible to simplify a subtraction which involves polynomials of different powers.

Ohhh...sorta got confused, but I knew it all along.

Thanks people.

Np....10th grade...how old are you?

15

[Clive]

27(3x^2-1)x^2y

Is what i get glancing at it..not sure if it's fully factorised/completed the right way...took the C2 module before the summer and already seem to have forgotten most the stuff...so much diff/int in FP and M3..

EDIT: You have a pretty crappy maths teacher if they can't tell you the answer to what they're supposed to be teaching you...

[Norse]

9X Log (2y + 2) = 27X Log (2y)

1/3 log(2y + 2) = Log (2y)

...

Howd you get an equals sign?

Btw, some advice, even if the first bit was right, the accepted answer would be "log(2y + 2) = 3Log (2y)". Always try to keep whole numbers, and avoid fractions, there very nasty.

Tit, thanks for the "advice". There ever so nasty. I know for a fact that GP will come here tommorow, and tell us that what you have provided him is not what the teacher was asking for...it's too simplistic.

I'm just trying to rack my brains 'twas a good few years since I have done pure maths...and I know this is solveable.

Jeez, you're fresh out of college...shouldnt you be able to regurgitate logarhtims like it was second nature to you?

[Clive]

Norse...that equation can't be solved it's not given as equal to anything...it's for simplifying only

[static_ice]

this is why I'm dropping out of Calc AB this year, I forgot everything, even something as easy as that

[Norse]

Norse...that equation can't be solved it's not given as equal to anything...it's for simplifying only

well, i was going on the f(x) principle...

but anyhoo


9({X^y}{X^2} - {3^Y})

Thats the lowest I can get it down to w/o using logs...

Dammit, I know theres a fooking simpler way of doing this...

[The Weird One]

Always try to keep whole numbers, and avoid fractions, there very nasty.

just depends on the teacher. my teacher right now hates decimals, so we have to answer everything in fractions. . . not that bad once you get used to them, really.

[Norse]

If you are unable to work using fractions, even extremely laborius ones, you shouldnt be doing maths.

[Skittles!]

Writing maths is so much easier than typing it.

[Titanic]

Norse...that equation can't be solved it's not given as equal to anything...it's for simplifying only

well, i was going on the f(x) principle...

but anyhoo


9({X^y}{X^2} - {3^Y})

Thats the lowest I can get it down to w/o using logs...

Dammit, I know theres a fooking simpler way of doing this...

Yer, the way I did it.

Remember hes 15, the solution is going to involve GCSE standard difficulty. No need for f(x) or logs. Its just the simple thing of finding the common product, removing it from the bracket, and leaving in what needs to be multiplied by the product to give the original answer. I have full confidence in the answer I gave to Greece.

Btw, what I was saying about fractions was not that I cant do them, but that I prefer working without them as do most A level maths students because with fractions its so easy to make a minor error, whilst with whole numbers its more straightforward.

[Norse]

To be honest though, for a 15 year old, that is some heavy mathematics.

I'm no mug when it comes to maths, but I never touched on algebra like that until 1st year a-level (16-17 years)

[Clive]

To be honest though, for a 15 year old, that is some heavy mathematics.

I'm no mug when it comes to maths, but I never touched on algebra like that until 1st year a-level (16-17 years)

Yeah same for me I didn't do that until I started 6th form.

[GreecePwns]

Actually, I did pretty good on the test, and much easier problems came up like factoring the difference of two sqares:

16x^4 - 144y^8k

(4x^2 + 12y^4k)(4x^2 - 12^4k)

I got a 92% on the test. Much better than expected.

And the accepted answer had nothing to do with logarithms or whatever you guys said (to be honest, i have never heard the word before, but did see a "log" button on my calculator).

(9x^[2y+2]) - (27x^2y) = 9x^2y (x^2 - 3)

P.S. This is an honors course, so it's like doing math from the year ahead.

[Kaplowitz]

im gettin scared! Im in 9th grade, Math BH, with a shitty teacher. But i its really easy proofs and stuff. I hope it never gets like this!

[GreecePwns]

:shock: im gettin scared! Im in 9th grade, Math BH, with a shitty teacher. But i its really easy proofs and stuff. I hope it never gets like this!

Ahh! I hate proofs with a passion!