The Remainder Theorem Help

The Remainder Theorem Help

Postby Guest » Thu Oct 18, 2012 12:42 pm

Hey guys, I am stuck on this question that I am revising for a test.
"The Remainder when x^3-2x^2+ax+5 is divided by x-3 is twice the remainder when the same expression is divided by x+1. Find the value of the constant a."

Could you please help me step-by-step? (the answer is 4 but I need to know how you get there).

Thanks in advance
Guest
 

Re: The Remainder Theorem Help

Postby Jonasmathwork » Wed Aug 21, 2013 2:33 am

Solution: Given that when x^3-2x²+ax+5 is divided by x-3 the reminder is 2

again when the same expression is devised by (x-1) then reminder is 4 ,what is the value of a=?

let a=-4
x^3-2x²-4x+5/(x-1)
the reminders are 6,2

let a=-3
x^3-2x²-3x+5/(x-1)
the reminders are 5,5

let a=-2
x^3-2x²-2x+5/(x-1)
the reminders are 8,4

there for the value of a=-2.

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Re: The Remainder Theorem Help

Postby arzoo » Wed Jan 22, 2014 8:40 am

Yeah I also got the solution!!!!!!! :D

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Re: The Remainder Theorem Help

Postby Guest » Wed Jan 22, 2014 12:04 pm

?????

Do you still need a solution over a year later...??????
Guest
 

Re: The Remainder Theorem Help

Postby Guest » Wed Jan 22, 2014 5:12 pm

The original question was about "The Remainder Theorem Help"

The question was about evaluating polynomials, factors, quotients, remainders etc and the relationships between them.

A polynomial P(x) when divided by (x-3) had twice the remainder than when the same polynomial was divided by (x+1).

If P(x) was divisable by (x-3) and the remainder was Zero then (x-3) would be a factor of P(x) and the expression would

have a Root or Zero at (x+3).....but not in our case here.
Similar for the other divisor (x+1), it would be a factor if remainder was Zero.....but is not zero here

writing the other way around...

P(x) = (x-3)x(Q(x)) + R1 and P(x) = (x+1)x(Q(x)) + R2

(x-3) is 0 when (x=3) and (x+1) is 0 when (x=-1)

So P(x) = R1 when x=3 and P(x) = R2 when x=-1

and we are told R1 = 2 x R2


So at the values (x=3) and (x=-1) the values of the polynomial P(x) is equal to the remainder.

P(x) = X^3 - 2X^2 + aX + 5


At (x=3)....

(3x3x3) + (-2x3x3) + (3a) + 5
= 27 - 19 + 3a +5
= (3a + 14)


At (x=-1)....

(-1x-1x-1) + (-2x-1x-1) + (-a) + 5
= -1 - 2 - a + 5
= (2 - a)


Now we are told that x=3 remainder is twice the x=-1 remainder so.....

2(2 - a) = (3a + 14)

4 - 2a = 3a + 14
4 - 14 = 3a + 2a
- 10 = 5a
- 2 = a

So....The value of the constant "a" is -2
---------------------------------------------

.........................Simple.....................
Guest
 

Re: The Remainder Theorem Help

Postby Guest » Sat Jan 25, 2014 12:02 pm

Small error in above posting........

" have a Root or Zero at (x+3).....but not in our case here. "
................................

The above should read..............
have a Root or Zero at ( x= (+3) ).....but not in our case here.

......................Simple........to make mistakes................
Guest
 

Re: The Remainder Theorem Help

Postby Guest » Fri Mar 18, 2016 1:53 pm

In these type of question,we should put the values of x for 3 and 1 and get remainder and find the value of a
A=2 is the correct answer
Guest
 


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