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# Maths geniuses, bit of help? C2 watch

1. It is given that p(x) = x^4 + x^2

a) Find the quotient and the remainder when p(x) is divided by (x-1)(x-3)

(Did that part and got it right, the quotient is: x^2 + 4x + 14
................................ .......the remainder is: 44x + 42

b) (This is where I have a problem):

Use the remainder theorem and your answer to part (a) to find the remainders (i) when p(x) is divided by 1 (ii) when p(x) is divided by 3.

????? Help would be appreciated..Thank you
2. Anyone?...or is it not just me?
3. Hey TwentyTen, are you sure youve posted that correctly, as isnt dividing p(x) by 1 or 3 quite trivial, ie

Ill assume you meant divided my (x-1) and (x-3). So consider, if I write the answer to your first part down :

now how to use the remainder theorm, well we know it is that the remainder of a polynomial p(x) devided my a linear divisor x-a is eqaul to p(a). Now one could simply use this staight away without doing the first part, but can you see it would be a little inconvinenet to evaluate the higher power x, but using the alternative form I have written above, you dont have to, hopefully you can have a go from there.

PS man youve got to be more patient, you cannot expect a reply right away, some people wait days for replys, if you impatient people simply wont help you
Hey TwentyTen, are you sure youve posted that correctly, as isnt dividing p(x) by 1 or 3 quite trivial, ie

Ill assume you meant divided my (x-1) and (x-3). So consider, if I write the answer to your first part down :

now how to use the remainder theorm, well we know it is that the remainder of a polynomial p(x) devided my a linear divisor x-a is eqaul to p(a). Now one could simply use this staight away without doing the first part, but can you see it would be a little inconvinenet to evaluate the higher power x, but using the alternative form I have written above, you dont have to, hopefully you can have a go from there.

PS man youve got to be more patient, you cannot expect a reply right away, some people wait days for replys, if you impatient people simply wont help you
Thanks for the help, but the book says "when p(x) is divided by 1" and "when p(x) is divided by 3"

(I used to always assume the book people got it wrong lol, not anymore )

....The answers at the back of the book say

a) 2
b) 90
...........................
5. (Original post by TwentyTen)
Anyone?...or is it not just me?
No, you are just very impatient as somebody didnt repsond to you in 14 minutes

As above, Im not sure you have written it down

Core 2 you will just have to evaluate linear factors) not a quadratic

so run the 2 arguments through the equation
6. Ok, right. Well thats got to be a missprint then or just a bad editing error as dividing an polynomial my constants such as 1 or 3 is a simple as its sounds. Secondly there was actually a mistake with your original remainer, I think I made the same mistake, its not 44x + 42, it should be 44x -42, not sure if that was simply a mistype or and maths error you made, but do go back and confirm that for yourself.

Now given that we have:

have a real think now, it is now simply a simple application of the remainder theorem and you get those answers
Ok, right. Well thats got to be a missprint then or just a bad editing error as dividing an polynomial my constants such as 1 or 3 is a simple as its sounds. Secondly there was actually a mistake with your original remainer, I think I made the same mistake, its not 44x + 42, it should be 44x -42, not sure if that was simply a mistype or and maths error you made, but do go back and confirm that for yourself.

Now given that we have:

have a real think now, it is now simply a simple application of the remainder theorem and you get those answers

Thank you! ......D.a.m.n misprints!!!

Multiplied it out and made x = 1 and x = 3
8. (Original post by dexter -1)
No, you are just very impatient as somebody didnt repsond to you in 14 minutes

As above, Im not sure you have written it down

Core 2 you will just have to evaluate linear factors) not a quadratic

so run the 2 arguments through the equation
Thanks for the help

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Updated: February 6, 2010
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