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

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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

Mathematics
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obviously Q(1) = 4 and Q(-1) = 8 by the remainder theorem
find the two values of x
x^2-1=(x-1)(x+1) so x=1 and x=-1 put one by one in equation

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so the reminder is 32
I dont follow that...
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
now get it ?
I don't think that's correct.
I'd dont think this is particularly different - I'm just missing something
*
O.o so what you think huh ? It is correct bro
so you just confused get calm and think on it for a minute
I dont' know - but I don't think your logic is correct
I'm going to look up the answer . I am confused - you are right there!! lol
its -4x + 9
yea! go an look the answer perhaps then you will believe on my answer
no way it can't be
take a screen short
I'm helping my grandson with his maths revision. Well that's the answer in the book.
I haven't got a scanner
it's very simple .. ok tell me how they get 4 when they divide equation by (x-1) huh ?
ok ganesh is here he can justify better now
4 is the remainder and = q(1)
Firstly, notice that we get a polynomial as remainder that is one degree less than whatever we're dividing by
^
right
for example, (x^5+2x+1)/(x^2-1) gives a remainder that looks like \(ax+b\) yes ?
right
similarly (x^100 + x+1)/(x^10 + 1) gives a remainder that looks like \(ax^9+bx^8+\cdots\)
the degree of remainder is always one less than the degree of bottom
so the remainder in this case must be of the form ax + b?
right, so lets suppose \[Q(x) = F(x)*(x^2-1)+\color{red}{ax+b}\] our goal is to find that red part
ok
since we know that \(Q(1)=4\) and \(Q(-1)=8\), plug them in and get two equations
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 ?
\[Q(1) = F(1)*(1^2-1)+\color{red}{a*1+b} \implies 4 = \color{red}{a+b} \tag{1}\] \[Q(-1) = F(-1)*((-1)^2-1)+\color{red}{a(-1)+b} \implies 8 = \color{red}{-a+b} \tag{2}\] two equations and two unknowns, we can solve them
for that, we may think that the coefficient of x is 0 @sohailiftikhar
thats really clever Thanx ganesh
np, im getting the remainder is \(-2x+6\) looks the options are wrong
yes i got that too b = 6 ans a = -2
I'll just recheck the answer in the book
lol
Yes thats the answer in the book. Well mistakes are made
happens... our method is pretty robust and straightforward, nothing that could go wrong..
from which grades book u got that problem bro ?
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...
lol ok
Open study is a great place to study . There is a wealth a talent here.
* wealth of talent
yes:)

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