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 March 18th, 2012, 01:46 PM #1 Newbie   Joined: Mar 2012 Posts: 17 Thanks: 0 synthetic division use synthetic division to show that (x+1/3)is a factor of g(x). than find factor to find the other two factors. list all three zeros g(x)=6x^3-x^2-4x-1
 March 18th, 2012, 03:38 PM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: synthetic division [color=#000000]$\begin{tabular}{c c c cccc| c } 6x^3 & - & x^2 &-& 4x&-& 1 & x+\frac{1}{3}\\\hline -6x^3 & -& 2x^2 &&&&& 6x^2-3x-3\\ &-&3x^2&-&4x&-&1& \\ &&3x^2&+& x&&&\\&&&-&3x&-&1&\\&&&&3x&+&1&\\&&&&&&\fbox{0}&\\ \end{tabular}$ $\hspace{175pt}\rule{1pt}{50pt}\\\hspace{175pt}\dot s\,\dots>\;residue\; zero$ Now for your other question $6x^3-x^2-4x-1=x^3-x^2+4x^3-4x+x^3-1=x^2(x-1)+4x(x^2-1)+(x-1)(x^2+x+1)=(x-1)(x^2+4x(x+1)+x^2+x+1)=$ $(x-1)(x^2+4x^2+4x+x^2+x+1)=(x-1)(6x^2+5x+1)=.......$ . [/color]
March 18th, 2012, 03:52 PM   #3
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Re: synthetic division

[attachment=0:3dfyrsve]synthdiv.jpg[/attachment:3dfyrsve]

So we have:

$g(x)=6x^3-x^2-4x-1=$$x+\frac{1}{3}$$$$6x^2-3x-3$$=$$3x+1$$$$2x^2-x-1$$=(3x+1)(2x+1)(x-1)$

The zeroes are therefore:

$x=-\frac{1}{2},\,-\frac{1}{3},\,1$
Attached Images
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 March 19th, 2012, 05:19 AM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: synthetic division [color=#000000]$\hspace{390pt}\fbox{\begin{tabular}{c |r r r r} &6& -1 & -4&-1 \\-1/3&& -2 & 1 & 1 \\\hline &6 &-3 & -3 &\rule{3pt}{16pt}\;0\\\end{tabular}}$ Mark's table with latex code (unfortunately some commands like \cline don't work). Code: \begin{tabular}{c |r r r r} %explaining it is the alignment(center + vertical line + right + right+right +right) &6& -1 & -4&-1 \\ -1/3&& -2 & 1 & 1 \\ \hline &6 &-3 & -3 &\rule{3pt}{16pt}\;0\\ \end{tabular} [/color]
 March 19th, 2012, 07:31 AM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: synthetic division Nicely done! I tried it with a slight modification... $\begin{tabular}{c |r r r r} &6& -1 & -4&-1 \\ \tiny{-\frac{1}{3}}&& -2 & 1 & 1 \\ \hline &6 &-3 & -3 &\rule{1pt}{18pt}\;0\\ \end{tabular}$
 March 19th, 2012, 07:40 AM #6 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: synthetic division [color=#000000]Seeing the modifications you did to my code, I sense you understood the logic behind it! The tabular environment works like a matrix![/color]

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