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

**1**

# 2002 Non Calc Winter Diet

Started by *Kiran*, May 18 2006 04:08 PM

9 replies to this topic

### #1

Posted 18 May 2006 - 04:08 PM

### #2

Posted 18 May 2006 - 04:10 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

### #3

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

you set your remainders for x=2 and x=-3 equal to zero and solve via simultaneous equations

edit: basically what SncZ said

### #4

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.

### #5

Posted 18 May 2006 - 04:30 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

### #6

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

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

### #7

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?

### #9

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...

what I had done was for the first bit (3) I densely put 0 at the bottom...

such a thicket...

### #10

Posted 18 May 2006 - 05:46 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.

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