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# A Few A Level Maths Questions watch

1. I have a few A Level maths questions which I can't seem to get my head around:

The polynomial f(x) is defined by f(x)=27x^3-9x + 2 .
(a) Find the remainder when f(x) is divided by 3x+1 .

The answer is apparently: 4 from getting -1+3+2.

To get that you must do 27(-1/3)^3-9(-1/3)+2

However, seeing as we are dividing by 3x I don't understand why X -1/3 as oppose to +1/3.

I'm only doing GCSE maths mind you but it seems like a big bloody step up. I'm taking AS maths a year early too, so I need to at least start understanding some of this crap now .
2. I'm crap at maths....but anyway

see f(x) as 3x+1=0

Take it all across you get [I'll do it step by step so you see whats happening]
3x+1=0
=>3x=-1
=>-1/3x

so f(x) =-1/3x

Then sub -1/3 in for x in the polynomial.

Don't ask why it is, it's just the way it is. Others may know, I just don't seeing as I'm expecting to fail my A2 in august
3. (Original post by Yousef)
I have a few A Level maths questions which I can't seem to get my head around:

The polynomial f(x) is defined by f(x)=27x^3-9x + 2 .
(a) Find the remainder when f(x) is divided by 3x+1 .

The answer is apparently: 4 from getting -1+3+2.

To get that you must do 27(-1/3)^3-9(-1/3)+2

However, seeing as we are dividing by 3x I don't understand why X -1/3 as oppose to +1/3.

I'm only doing GCSE maths mind you but it seems like a big bloody step up. I'm taking AS maths a year early too, so I need to at least start understanding some of this crap now .
The remainder theorem is that if you divide by then the remainder is
Basically you shove into the expression what ever you get for x when the thing you are diving by equals 0.

So 3x+1 = 0 and then x= -1/3
4. That how i did it for c1 ^^ i always = it to 0 and then solve it ... substitute x into the eq and get the right answer.... what more can you ask for
5. f(x)/(3x+1) = p(x) with remainder R(x) (where p(x) is some unknown polynomial)

then f(x) = p(x)(3x+1) + R(x)

setting x = -1/3 makes (3x+1) = 0

and so f(-1/3) = R(-1/3) which is the remainder
6. (Original post by Loz17)
Don't ask why it is, it's just the way it is.
Here's why. What you're trying to do is find , where . If you multiply through by , then you get . Notice that if you let , then on the RHS you're just left with , and the LHS becomes a numerical value; this is the remainder.
7. I miss C1

I hate C4
8. (Original post by nuodai)
Here's why. What you're trying to do is find , where . If you multiply through by , then you get . Notice that if you let , then on the RHS you're just left with , and the LHS becomes a numerical value; this is the remainder.
Ahhh

Thank you

I would have thought 27x^3/9x^2 = 3x

-9x/3x = -3

2/-2 = -1

Which would be 3x-4, however the answer is meant to be 3x+1.
10. You can be a sweet dream

or a beautiful nightmare

Either waY I Dont want to wake up from you
11. (Original post by Yousef)

I would have thought 27x^3/9x^2 = 3x

-9x/3x = -3

2/-2 = -1

Which would be 3x-4, however the answer is meant to be 3x+1.
You can't "cancel" things like that. (Is (1+2)/(1+8) the same as 1/1 + 2/8?)

Cancellation (convince yourself of this with a numerical example if you like) is dividing the top and bottom of a fraction by common factors. For example, if you wanted to simplify , you could factorise (factor-ise, geddit?) the top into (x+1)(x+2), notice that we also have an x+1 on the bottom, and cancel those. Now, you're expected to do the same with your slightly uglier fraction. I recommend you factorise the denominator first, because that's quadratic and the other one's cubic, and then check whether the numerator has any of those factors.
12. Ah, that makes much more sense thanks .

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