I guess i've done problems like this in the past but I'm struggling with this one:-
The polynomial Q(x) leaves remainder 4 when divided by x - 1, and remainder 8 when divided by x + 1. The remainder when Q(x) is divided by x^2 - 1 is
A 32
B -4x + 9
C -4x - 7

- welshfella

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- welshfella

obviously Q(1) = 4 and Q(-1) = 8 by the remainder theorem

- sohailiftikhar

find the two values of x

- sohailiftikhar

x^2-1=(x-1)(x+1)
so x=1
and x=-1 put one by one in equation

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## More answers

- sohailiftikhar

so the reminder is 32

- welshfella

I dont follow that...

- sohailiftikhar

as given in data reminder for (x-1)=4 and for (x+1)=8
so for x^2-1=(x-1)(x+1)=4*8=32

- sohailiftikhar

now get it ?

- welshfella

I don't think that's correct.

- welshfella

I'd dont think this is particularly different - I'm just missing something

- geerky42

*

- sohailiftikhar

O.o so what you think huh ? It is correct bro

- sohailiftikhar

so you just confused get calm and think on it for a minute

- welshfella

I dont' know - but I don't think your logic is correct

- welshfella

I'm going to look up the answer . I am confused - you are right there!! lol

- welshfella

its -4x + 9

- sohailiftikhar

yea! go an look the answer perhaps then you will believe on my answer

- sohailiftikhar

no way it can't be

- sohailiftikhar

take a screen short

- welshfella

I'm helping my grandson with his maths revision. Well that's the answer in the book.

- welshfella

I haven't got a scanner

- sohailiftikhar

it's very simple .. ok tell me how they get 4 when they divide equation by (x-1) huh ?

- sohailiftikhar

ok ganesh is here he can justify better now

- welshfella

4 is the remainder and = q(1)

- ganeshie8

Firstly, notice that we get a polynomial as remainder that is one degree less than whatever we're dividing by

- Nnesha

^

- welshfella

right

- ganeshie8

for example, (x^5+2x+1)/(x^2-1) gives a remainder that looks like \(ax+b\) yes ?

- welshfella

right

- ganeshie8

similarly (x^100 + x+1)/(x^10 + 1) gives a remainder that looks like \(ax^9+bx^8+\cdots\)

- ganeshie8

the degree of remainder is always one less than the degree of bottom

- welshfella

so the remainder in this case must be of the form ax + b?

- ganeshie8

right, so lets suppose
\[Q(x) = F(x)*(x^2-1)+\color{red}{ax+b}\]
our goal is to find that red part

- welshfella

ok

- ganeshie8

since we know that \(Q(1)=4\) and \(Q(-1)=8\), plug them in and get two equations

- sohailiftikhar

ganesh what you said about the given that reminder of that equation is 4 when divided by (x-1) where is x term with 4 ?

- ganeshie8

\[Q(1) = F(1)*(1^2-1)+\color{red}{a*1+b} \implies 4 = \color{red}{a+b} \tag{1}\]
\[Q(-1) = F(-1)*((-1)^2-1)+\color{red}{a(-1)+b} \implies 8 = \color{red}{-a+b} \tag{2}\]
two equations and two unknowns, we can solve them

- ganeshie8

for that, we may think that the coefficient of x is 0 @sohailiftikhar

- welshfella

thats really clever Thanx ganesh

- ganeshie8

np, im getting the remainder is \(-2x+6\)
looks the options are wrong

- welshfella

yes i got that too
b = 6 ans a = -2

- welshfella

I'll just recheck the answer in the book

- sohailiftikhar

lol

- welshfella

Yes thats the answer in the book. Well mistakes are made

- ganeshie8

happens... our method is pretty robust and straightforward, nothing that could go wrong..

- sohailiftikhar

from which grades book u got that problem bro ?

- welshfella

Oh its a pretty old UK Advanced Level book from 1979. Examinations have become a little easier since then. Its wriiten by a professor of Mathematics but mistakes are made by everyone...

- sohailiftikhar

lol ok

- welshfella

Open study is a great place to study . There is a wealth a talent here.

- welshfella

* wealth of talent

- sohailiftikhar

yes:)

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