Posted 18 May 2006 - 04:08 PM
Hey Guys. This is probably really easy but I was wondering if anyone could help me with this as I am quite stuck.
"Given that (x - 2) and (x+3) are factors of f(x) where f(x) = 3x[^3] + 2x[^2] + cx +d, find the values of c and d."
I thought it may have something to do with (x-2)x(x+3) but I can't seem to crack it....
The answers are c = -19, d=6
Thanks in advance guys
Posted 18 May 2006 - 04:10 PM
QUOTE(*Kiran* @ May 18 2006, 05:08 PM)

Hey Guys. This is probably really easy but I was wondering if anyone could help me with this as I am quite stuck.
"Given that (x - 2) and (x+3) are factors of f(x) where f(x) = 3x[^3] + 2x[^2] + cx +d, find the values of c and d."
I thought it may have something to do with (x-2)x(x+3) but I can't seem to crack it....
The answers are c = -19, d=6
Thanks in advance guys

you set up synthetic division and divide by (x-2) and (x+3) . since they're factors that you know they equal zero.

you just set up simultaneous equations and solve for c & d
Posted 18 May 2006 - 04:14 PM
use synthetic divsion..or other method to factorise it
you set your remainders for x=2 and x=-3 equal to zero and solve via simultaneous equations

edit: basically what SncZ said
Posted 18 May 2006 - 04:25 PM
You don't have to do he simultaneous equation if you do synthetic division with (x-2) and then do more synthetic division with the bigger bracket (which only has c in it) Then you just find c and sub in to get d.
Posted 18 May 2006 - 04:30 PM
QUOTE(ice_illusion @ May 18 2006, 05:25 PM)

You don't have to do he simultaneous equation if you do synthetic division with (x-2) and then do more synthetic division with the bigger bracket (which only has c in it) Then you just find c and sub in to get d.
um not sure what you mean, i just done it and all my equations and brackets have c and d in them.
the general rule is: if you have 2 unknown variables, you need 2 equations to solve it.
and if u have 3 unknown variables you need 3 equations etc etc
Posted 18 May 2006 - 04:39 PM
thanks guys...but I'm still kinda having problems...
once i have synthetically divided it i end up with :
8 + 2c +d for (x-2)
and
-6+(-3c) +d for (x+3)...
when I use simultaneous equations i end up with really weid answers... so I wonder if someone could please help me.. i am so dense . (I really am trying btw...)
much appreciated
Posted 18 May 2006 - 04:44 PM
I think you've made a slip in the synthetic division - have you remembered to multply by 2 (or 3) each step?
Posted 18 May 2006 - 05:00 PM
QUOTE(George @ May 18 2006, 05:44 PM)

I think you've made a slip in the synthetic division - have you remembered to multply by 2 (or 3) each step?

george im jealous of your magic powers

how do you draw up a synthetic division table ?
Posted 18 May 2006 - 05:02 PM
Waw! thanks very much guys...
what I had done was for the first bit (3) I densely put 0 at the bottom...
such a thicket...
Posted 18 May 2006 - 05:46 PM
QUOTE(SncZ @ May 18 2006, 06:00 PM)

george im jealous of your magic powers

how do you draw up a synthetic division table ?

I used
this program to make up the latex code, then pasted it in [tex] tags - you can see the code by quoting my post.