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2002 Non Calc Winter Diet


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#1 *Kiran*

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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 thumbsup.gif

#2 SncZ

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Posted 18 May 2006 - 04:10 PM

QUOTE(*Kiran* @ May 18 2006, 05:08 PM) View Post

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 thumbsup.gif



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

therefore.gif you just set up simultaneous equations and solve for c & d

#3 Nathan

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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 biggrin.gif

edit: basically what SncZ said wink.gif

#4 ice_illusion

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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 SncZ

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Posted 18 May 2006 - 04:30 PM

QUOTE(ice_illusion @ May 18 2006, 05:25 PM) View Post

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 *Kiran*

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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 smile.gif

#7 George

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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?

\begin{tabular}{c|cccc}
2 & 3 & 2 & c & d \\ 
 &  & 6 & 16 & 32+2c \\ 
\cline{2-5}
\multicolumn{1}{c}{} & 3 & 8 & 16+c & 32+2c+d \\ 
\end{tabular}

#8 SncZ

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Posted 18 May 2006 - 05:00 PM

QUOTE(George @ May 18 2006, 05:44 PM) View Post

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

\begin{tabular}{c|cccc}
2 & 3 & 2 & c & d \\ 
 &  & 6 & 16 & 32+2c \\ 
\cline{2-5}
\multicolumn{1}{c}{} & 3 & 8 & 16+c & 32+2c+d \\ 
\end{tabular}



george im jealous of your magic powers sad.gif how do you draw up a synthetic division table ?

#9 *Kiran*

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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...

#10 George

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Posted 18 May 2006 - 05:46 PM

QUOTE(SncZ @ May 18 2006, 06:00 PM) View Post

george im jealous of your magic powers sad.gif how do you draw up a synthetic division table ?

laugh.gif 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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