Question

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

Answer 1

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

Answer 2

find the two values of x

Answer 3

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

Answer 4

so the reminder is 32

Answer 5

I dont follow that...

Answer 6

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

Answer 7

now get it ?

Answer 8

I don't think that's correct.

Answer 9

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

Answer 10

*

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

its -4x + 9

Answer 16

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

Answer 17

no way it can't be

Answer 18

take a screen short

Answer 19

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

Answer 20

I haven't got a scanner

Answer 21

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

Answer 22

ok ganesh is here he can justify better now

Answer 23

4 is the remainder and = q(1)

Answer 24

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

Answer 25

^

Answer 26

right

Answer 27

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

Answer 28

right

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

right, so lets suppose
\[Q(x) = F(x)*(x^2-1)+\color{red}{ax+b}\]

our goal is to find that red part

Answer 33

ok

Answer 34

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

Answer 35

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 ?

Answer 36

\[Q(1) = F(1)*(1^2-1)+\color{red}{a*1+b} \implies 4 = \color{red}{a+b} \tag{1}\]

\[Q(-1) = F(-1)*((-1)^2-1)+\color{red}{a(-1)+b} \implies 8 = \color{red}{-a+b} \tag{2}\]


two equations and two unknowns, we can solve them

Answer 37

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

Answer 38

thats really clever Thanx ganesh

Answer 39

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

Answer 40

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

Answer 41

I'll just recheck the answer in the book

Answer 42

lol

Answer 43

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

Answer 44

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

Answer 45

from which grades book u got that problem bro ?

Answer 46

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...

Answer 47

lol ok

Answer 48

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

Answer 49

* wealth of talent

Answer 50

yes:)