**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** :

http://getrevising.co.uk/resources/maths_question_worked_example

**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 ::

http://getrevising.co.uk/resources/lucia_your_answer

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

BraniacEdited by

Braniac on Fri 22nd June, 2012 @ 14:46