Edexcel Maths C3

I hope c4 will be horrendously hard. I hope Edexcel will have enough backbone to add some STEP/AEA level questions in there to compensate for the lack, sorry i meant the non existence of hard questions in c3 while keeping the grade boundaries high. We can't have everyone running around with A*s.

I hope c4 will be horrendously hard. I hope Edexcel will have enough backbone to add some STEP/AEA level questions in there to compensate for the lack, sorry i meant the non existence of hard questions in c3 while keeping the grade boundaries high. We can't have everyone running around with A*s.

nope, i used remainder theorem, i divided the fraction obtained a remainder, did the answer of divison plus the remainder over the divisor

so bassically if you make my answer with a common denominator i get what u got.

they can not add questions that are not on the specification. How easier or hard the paper is perceived to be is compensated by the fact the boundaries will change.

However, I would like either C3 or C4 to be on the hard side because it will then allow very able students to make non-conceptual errors and still get an A*

I am sorry I just not sure what you mean; I am not sure you understand what the remainder theorem is.

if you divide the denomination into the numerator you get a quotient of 1 and a remainder of x^2 + 7x - 10.

2x^2 + 9x -5 divided by x^2 + 2x -15 = 2 with a remainder of 5x+25

according to the first chapter in c3 new book:

f (x) = 2 + 5x+25/ x^2 + 2x -15

5x + 25 = 5( x+5)

factorise x^2 + 2x -15 you will get (x +5) ( x-3)

cancel x+ 5

you get 5/ x-3 + 2


now, if you make this with same denominator i get ur answer which 2x-1/x-3

ah, with you. This doesn't have anything to do with the quotient rule or the remainder theorem by the way.

Well technically, what you have done is correct and it doesn't say to express as a single fraction so I think you will get the marks.

I would say you will lose a method and accuracy mark.

Arsey deserves a bloody medal for their input on this thread...

Thanks for the ms.

On question 7c, I can't remember the answers I put but I think I may have got them right. I found dH/dt and set it equal to 0, found t and substituted that into the original equation to find H. Assuming I did the method properly, would it work?