please help!! x^3 + ax^2 + bx+ c

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

1 Attachment(s)

Re: please help!! x^3 + ax^2 + bx+ c

Attachment 23314

please right the way of solving it

Re: please help!! x^3 + ax^2 + bx+ c

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.

Re: please help!! x^3 + ax^2 + bx+ c

by (x-a) you mean (x-1) ?!

because i didnt get it

and no i dont have more information

Re: please help!! x^3 + ax^2 + bx+ c

wait ... i got it

but how did the 3 appeared ?!

Re: please help!! x^3 + ax^2 + bx+ c

can u explain more please

Re: please help!! x^3 + ax^2 + bx+ c

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.

Re: please help!! x^3 + ax^2 + bx+ c

and the b=3/2 ?!

because if i only know this point

i will solve it

Re: please help!! x^3 + ax^2 + bx+ c

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

??

Re: please help!! x^3 + ax^2 + bx+ c

i've solved it :D

thank you

u saved my life

Re: please help!! x^3 + ax^2 + bx+ c

Re: please help!! x^3 + ax^2 + bx+ c

I'm glad I was of help.

Quote:

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.

Quote:

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.