Hi
so I am a student of Maths and Physics Track in my school
And in todays class in Maths, the teacher gave us a question. And when no one could solve it, he said that there is going to be a quiz in it tomorrow
I tried to solve it several times but I couldn't solve it

anyways,, the question is

if :: x^3 + ax^2 + bx + c
the rest of dividing it on (x-1) is :: 4
the rest of dividing it on (x+1) is :: -1

find :: a , b , c

please right the way of solving it

The remainder when polynomial f(x) is divided by x-a is f(a). Considering this you get a+b+c = 3 abd a-b+c = 0. So b = 3/2 and a+c = 3/2. You can only have the solution in that form unless you have more information about the concerned polynomial.

by (x-a) you mean (x-1) ?!
because i didnt get it

wait ... i got it
but how did the 3 appeared ?!

The remark of the remainder of f(x) being f(a) when divided by x-a is the statement of the remainder theorem, and the letter a is not the coefficient a of your problem here. The 3 appeared because f(1) = 4 implies 1+a+b+c = 4 which implies a+b+c = 3.

and the b=3/2 ?!
because if i only know this point
i will solve it

do u mean
when we have a positive B and a negative B
we count them as 2B
and to find B
we make this 0+3
so it will be like
2B = 3
B = 3/2
??

i've solved it
thank you
u saved my life

a = -2.5
b = 1.5
c = 4

I'm glad I was of help.

Originally Posted by Multi
do u mean
when we have a positive B and a negative B
we count them as 2B
and to find B
we make this 0+3
so it will be like
2B = 3
B = 3/2
??
Yes, you subtract the two equations from one another:

a+b+c = 3
a-b+c = 0
----------
2b = 3

=> b = 3/2.

a = -2.5
b = 1.5
c = 4
Yes, that works, but so does (a, b, c) = (1, 3/2, 1/2) and many/infinite others. In fact, any a, c ∈ ℝ such that a+c = 3/2 works.
I would leave it in the general form unless the teacher asked only for a pair or you have more restrictions on the given polynomial.