# Thread: find missing integer-algebraic division

1. ## 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

2. 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

4. ## got it

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
cool!
thank you

5. 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

6. ## re: why?

why do you do this Skeeter?

7. 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?

8. Originally Posted by lemonz
why do you do this Skeeter?
good question ... it's probably a form of insanity.

9. ## re: nuttz!

well, thanks crazy person

10. 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

11. ## re: synthetic division check

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

12. x^3+bx^2-15x+3 divided by x+1 remainder 14: Find b
quick check ... see if $\displaystyle f(-1) = 14$ when $\displaystyle b = -3$

$\displaystyle f(x) = x^3-3x^2-15x+3$

$\displaystyle f(-1) = \, ?$