beelal I need your help!:)

more maths questions!

1) find the remainder when 6x (cubed) - 13 x (squared) + x + 2 is divided by x+3

2) show that x-2 is a factor of 6x (cubed) - 13 x (squared) + x + 2

3) hence factorise 6x (cubed) - 13 x (squared) + x + 2

4) The coordinates of the points A and B are (2,8) and (4, -6) respectively

calculate the length of line AB

5) Find the equation of the straight line perpendicular to AB that passes through the midpoint of AB

Posted Fri 22nd June, 2012 @ 13:50 by AmyPond
Edited by AmyPond on Sat 23rd June, 2012 @ 18:21

Oh yh my little x's mean multiply and my big X means that it is a consant!!

Well question 1 is the following!! :

Basically to work out the remainder quickly do this :

f (x) = 6X^3  -  13X^2   +  X  +  2

Now u can sub in -3 as x because you can rearrange  X  + 3 = 0  to make X = -3

Now sub that into the equation and you get ::

=[6x(-3)^3]  -  [13x(-3)^2] + (-3) + 2

this equals out to ===> -280

Now for questions 2 you can use the same method and basically all you have to do is sub X=2 and then you should get zero as your answer because it says that it is a factor !!

X = 2

sooo

y  =  [6*(2*2*2)] -[13*(2*2)] +2 + 2

=  0

Now for question 3 you have to do division of polynomials :

Basically for question 4 you need to use pythagoras!!

So draw a quick sketch and you will get two points !! now what you can do is draw a line linking A and B together!!

Now what you do is you need another point to make a triangle so have a look at this picture ::

as you can see on my drawing that there is a point C and that will form a right angle traingle allowing you to do pythagoras!!

So the values you get for the change in Y is 14 and the X is 2

so do AB (squared) = Y^2    +  X^2

=  (14 x 14) + (2 x 2)

= 200

AB   = Sqaure root (200)

= 14.14

Well for question 5 you have to work out the midpoint of AB then, do the reciprical gradient of AB then work out the y-intercept then you have your answer ... let me show u ;D xxxx

Midpoint is  (2+4) /2              (8-6)/2

3                      1

Midpoint is (3,1)

Gradient of AB = change in y                 14

------------    =     -----------  = 7

change in x                  2

Perpendicular gradient =   1

----

7

Now to work out the equation by subbing in your midpoint !!

y = 1/7 X + c

c  = y - (m x X)

= 1 - (1/7 x 3)

=  4/7

So your final equation is :::

Y = 1/7 X  +  4/7  or           X - 7Y +4 = 0

hope i helped ;D xxxx

Answered Fri 22nd June, 2012 @ 14:41 by Braniac
Edited by Braniac on Fri 22nd June, 2012 @ 14:46

okay basically!! if you look at my wokings out which i did, the top is what is left over!! Basically you have to say well what do i have to times x+2 by to get 6X^3  which is 6X^2 etc etc

Then u multiply x+2 by the 6X^2 and then you minus it from the first two parts of the equation 6x^3 - 13x^2 and then you will get a value!!

U then drop the next value down which is x n then the process continues until you reach zero

sorry for the quick reply ill be going out in a min so if i have time ill make it  better soz ***

Answered Fri 22nd June, 2012 @ 19:15 by Braniac

Well i culd have but i saved you time and this getd you the full marks as well ;D xxxx

Answered Fri 22nd June, 2012 @ 20:40 by Braniac

lol maybe i was wrong :/ it is possible u r cleverer than me ;D xxxx

Answered Fri 22nd June, 2012 @ 21:06 by Braniac

yh lucia if im on il try and answer them bbez dont u worry ;D xxxx

Answered Sat 23rd June, 2012 @ 08:01 by Braniac

sorry lucia can i answer this tomorrow but ill copy the question down and do it n then answer it tomorrow!!

Then i can just copy it out !!!

Soz again but yh i know how to do it

Answered Sat 23rd June, 2012 @ 21:06 by Braniac

kk nps bbez ;D ***

N ur  smarter than me ;) ***

Answered Sat 23rd June, 2012 @ 23:24 by Braniac

well for the extra question wot i think u can do is work out a straight line for two points and then  sub in the third x point and see if it crosses then it will have the same y value! That is my thought for just lookin at the question have a go at that bbez ***

Answered Sun 24th June, 2012 @ 16:41 by Braniac

yh lucia im in the process of workin it out bbez n i might post the answer tmoz or today if that is okay!!! soz have been sick all day bbez :(

Answered Sun 24th June, 2012 @ 20:21 by Braniac

ill post it b4 bbez in the morning coz i have a free period !!! ***

Answered Sun 24th June, 2012 @ 20:32 by Braniac

Ill give u wot i have at the moment !!

I did 2*(x+1)  + 2*(y+3) = 62

2x+2y = 54

Then u expand n make y the subject !!!

so u get i think y = 27-x

then u sub it in to the equation:

(x+9)(2x+y)=703

This gets you (x+9)(x+27) = 703

then u expand n get a quadratic !!!

And  you get X^2  +  36x  +  243 = 703

Make the equation equal zero !!

So you get :               X^2  +  36x  -  460  =  0

(x+46)(x-10) = 0

x = 10

x=-46

Sub the x value in one at a time so sub in to this equation as it is easier !!   ===> 2x+2y = 54

so when x = 10 :

(2*10 )  + 2y = 54

y = 17

so when x  -46

(2*-46) + 2y = 54

y = 73

Now u cannot have a negative x becaus e if you look on one of the sides it is (x+1) so if you had subbed -46 then you would have a side which is -45  which is impossible !!!!!!

Therefore this leaves you  with the answer that  X = 10   and y = 17 !!

Now  you can sub those values in and then you will get the dimensions of both rectangles!

Sorry if i was a bit late !!

Answered Sun 24th June, 2012 @ 21:03 by Braniac
Edited by Braniac on Mon 25th June, 2012 @ 09:41

Now for question 3, it really is simple indices !!

Okay write out the equation like so :

6X^3                       5 X^1/4

-----------      x       ------------

2                               (x^5)^1/4

Now what you can do is take the numbers to the side of the euation n just leave the X's on the top  of the fraction !! You can do this because you are just multiplying the number by the constant X so by keeping it to the side will make it easier to see what i am going to show you =D

Okay so keep it to the side (i mean the numbers !! ) :

X^3                                x^1/4

--------      x       5   ------------

2                                (x^5)^1/4

As you can see thaht the 2 can cancel out with the 6 to make 3 !! Then you can multiply the 3 by the 5 which u also took out leaving you with this so far:

X^1/4

15X^3    x     -----------------

(x^5)^1/4

Now u use the rule of indices which is when you do (X^a)^b  , you multiply the powers together !! so that is what you do at the denominator and you should get this:

X^1/4

15X^3    x     -----------

x^5/4

Now you also know that when you divide powers you minus them together so you should get this:

15X^3    x    ( X^1/4-5/4)

This gives you  15X^3   x   X^-1

Now for the finale bbez ;D xxxx

Wot you can do next is that you know the rule indices state that if you multiply powers you add them, so you get :

15X^ 3-1

Which finally gives you  ==> 15X^2

Hope i helped and sorry again for being soo late :'(

Answered Mon 25th June, 2012 @ 09:56 by Braniac

soz didnt get time to answer the other 1 !!!!

reallly sorry but if i get more time ill post the answer but ive gotta go lesson now byeeeeee !! ;D ***

Answered Mon 25th June, 2012 @ 09:58 by Braniac

thank you so much:) youve helped me such a lot!

um.. sorry to trouble you but do you think you can have a go at explaining the poly-nomal thingy again?

not sure how you got 6x^2- x - 1?

Answered Fri 22nd June, 2012 @ 14:49 by AmyPond
Edited by AmyPond on Fri 22nd June, 2012 @ 14:54

no its ok ive got it! um, though, why didnt you do the polynomials thing with q1? just curious, as it says find the remainder and divde and stuff.. not saying youre wrong and its only a 2 mark question on the paper... but just wondering:):)

Answered Fri 22nd June, 2012 @ 19:55 by AmyPond

thank you:) i did try the long method for q1 and got 96 as a remainder not -280....did i go really wrong?!

Answered Fri 22nd June, 2012 @ 20:46 by AmyPond

No way you are really smart and obviously know what you're talking about:) but I got the same as you using the x=-3 thing so maybe I've done it wrong on the long equation:-) can I ask you for help on other questions I'm stuck on? Those were only a few! Thank you lots!!:):)

Answered Fri 22nd June, 2012 @ 21:36 by AmyPond

hey... if ure doin lang will u be able 2 answer my discussion ,thankx

Answered Fri 22nd June, 2012 @ 21:38 by SciTech

Beelal- think I know where I went wrong:-) I'm learning! Just slowly...

Answered Fri 22nd June, 2012 @ 22:00 by AmyPond

Thanks beelal:) if this gets annoying just say..:)

1) A rectangle A has sides of length (x+1) cm and (y+3) cm.

The perimeter of A is 62cm.

A rectangle B has the sides of length (x+9) cm and (2x+ y) cm

The area of b is 703 cm (squared)

Calculate the dimensions of both these rectangles.

2) find the x - coordinate on the point of the curve y= 2x (squared) + 2x where the tangent to the curve is 12...

3) simplify 6x^3/2 X 5x^1/4 DIVIDED BY (x^5) ^1/4

Thank you:):):)

Answered Sat 23rd June, 2012 @ 08:53 by AmyPond
Edited by AmyPond on Sat 23rd June, 2012 @ 18:21

Beelal! I need your help again! sorry!

Answered Sat 23rd June, 2012 @ 18:17 by AmyPond

Yeah no problem:-) thank you soo much!

Answered Sat 23rd June, 2012 @ 21:32 by AmyPond

this is an extra - doing loads of past papers and keep finding new questions that i cant do!

is it possible to conect the coordinates (-1, 2) (8, 47) and (-10, -43) with a single straight line? you must show working and state a reason for your answer.

forever grateful!:)

Answered Sun 24th June, 2012 @ 16:36 by AmyPond

How do you work out a straight line? And I don't mind if you dot have time for all the questions but do you think, if you only did one- that you could look at the first one about rectangles please? That comes up on a lot of past papers..thank you:):)

Answered Sun 24th June, 2012 @ 17:27 by AmyPond

Ah no worries then:) sorry you're ill:/ my exam is at 9.00 am tomorrow so don't worry if you can't post it- I'll wing it;)

Answered Sun 24th June, 2012 @ 20:29 by AmyPond

Don't worry- I leave the house at 8 and don't have Internet access! I think the rectangle one is a simultaneous equation but I just can't figure out how to get those two equations out...:)

Answered Sun 24th June, 2012 @ 20:51 by AmyPond

No that's great:) thank you! I was trying to solve them independently and I get it now:):) thank you!!!!!!

Answered Sun 24th June, 2012 @ 21:23 by AmyPond

Thank you:)

Answered Mon 25th June, 2012 @ 18:24 by AmyPond