document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=366845>Question 366845</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If (2x-1) is a factor of f(x)=2x^3-5x^2-kx+3 find the value of k.\r<BR>\n" );
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document.write( "Find the other two factors of f(x)=0 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #261716 by </B><A HREF=/tutors/aboutme.mpl?userid=Jk22><B>Jk22(389)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Jk22><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Jk22><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=261716\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Dividing the polynomials gives : <BR>\n" );
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document.write( "2x^3 - 5x^2 - kx + 3  | 2x - 1<BR>\n" );
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document.write( "2x^3 - x^2............| x^2 - 2x - (k+2)/2<BR>\n" );
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document.write( "-4x^2 - kx + 3........|<BR>\n" );
document.write( "-4x^2 + 2x............|<BR>\n" );
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document.write( "-(k+2)x + 3...........|<BR>\n" );
document.write( "-(k+2)x + (k+2)/2.....|<BR>\n" );
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document.write( "3 - (k+2)/2<BR>\n" );
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document.write( "the remainder should equals zero, hence k+2 = 6, k = 4<BR>\n" );
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document.write( "we can check another way by the fact the 1/2 has to be a solution : <BR>\n" );
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document.write( "2/8 - 5/4 - 2 + 3 = 1/4 - 5/4 + 1 = -4/4 + 1 = 0<BR>\n" );
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document.write( "the remaining polynomial is : x^2 - 2x - 3 = (x-3)(x+1) by Viete's formula.<BR>\n" );
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document.write( "the 2 other roots are : 3 and -1<BR>\n" );
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document.write( "check : for x=3 : 2*27 - 5*9 - 4*3 + 3 = 54 - 45 - 12 + 3 = 9 - 12 + 3 = 0<BR>\n" );
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document.write( "x=-1 : -2 - 5 + 4 + 3 = -7 + 7 = 0<BR>\n" );
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