Thread: find missing integer-algebraic division

Hi:
I have come across a question but have not been able to work out a method to solve this, yet
When x^3+3x^2+ax+2 is divided by x – 2 the remainder is 4. Find a.

Any help. thanks
err - most help. thanks

use synthetic division ... you know the remainder is 4

Code:

2].......1........3........a.........2
..................2.......10.......20+2a
------------------------------------------
.........1........5.....10+a.......22+2a

22+2a = 4

hmmm!
don't know synthetic division. Is that high-school level?
I think I am only supposed to be doing algebraic division

hmmm!
think i got it
bring down the leading coefficient to the bottom row
multiply c by the value just written on the bottom row
add the column created in last step
repeat until done
write answer out and equate
cool!
thank you

Originally Posted by lemonz

hmmm!
don't know synthetic division. Is that high-school level?
I think I am only supposed to be doing algebraic division

most definitely high school ...

Synthetic Division

why do you do this Skeeter?

Originally Posted by lemonz

Hi:
I have come across a question but have not been able to work out a method to solve this, yet
When x^3+3x^2+ax+2 is divided by x – 2 the remainder is 4. Find a.

Any help. thanks
err - most help. thanks

Basic theorem: if P(x) is a polynomial and dividing by x-b gives quotient Q(x) with remainder R (which is a constant, a number) then P(x)/(x-b)= Q(x)+ R/(x-b) so P(x)= Q(x)(x- b)+ R. Setting x= b, P(b)= Q(b)(b-b)+ R= R no matter what Q(b) is. That is, the remainder when P(x) is divided by x- b is the value of P(b).

Here, P(2)= 2^4+ 3(2^2)+ a(2)+ 2= 16+ 3(4)+ 2a+ 2= 2a+ 30. What value of a makes that equal to 4?

Originally Posted by lemonz

why do you do this Skeeter?

good question ... it's probably a form of insanity.

well, thanks crazy person

So, is this right?

x^3+bx^2-15x+3 divided by x+1 remainder 14: Find b

1: 1 b -15 3
1 1+b -14+b
1 1+b -14+b 17+b

17+b=14
b=-3

If this is so, it mean that in the case of the last column, where -14 is under the 3, I ignore the minus sign and just make addition with integer.
But in the case of the 3rd column, I can't ignore the minus sign on the 15, cause it is the starting number?
thanks

So, is this right?

x^3+bx^2-15x+3 divided by x+1 remainder 14: Find b

1:::::1 b -15 3

1 1+b -14+b

1 1+b -14+b 17+b

17+b=14
b=-3

If this is so, it mean that in the case of the last column, where -14 is under the 3, I ignore the minus sign and just make addition with integer.
But in the case of the 3rd column, I can't ignore the minus sign on the 15, cause it is the starting number?
thanks

x^3+bx^2-15x+3 divided by x+1 remainder 14: Find b

quick check ... see if when