document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1128995>Question 1128995</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Is 2x(5x^3+6x^2-3x) completely factored. Why or why not. If not how do i factor the polynomial completely.\r<BR>\n" );
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document.write( "Thank you for your help\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #745528 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=745528\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1): Is 2x(5x^3+6x^2-3x) completely factored. Why or why not. If not how do i factor the polynomial completely. <BR>\n" );
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document.write( "Answer:: No because the 2nd factor can be furter factored::<BR>\n" );
document.write( "2x(5x^3+6x^2-3x) = 2x[x(5x^2+6x-3)] = 2x^2[5x^2+6x-3]<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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