document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=582099>Question 582099</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Does anyone have a easy way to factor?  Her is a problem I just don't know how to do.\r<BR>\n" );
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document.write( "9ax^3+15ax^2-14ax </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #371937 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=371937\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Does anyone have a easy way to factor?  Her is a problem I just don't know how to do.\r<BR>\n" );
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document.write( "9ax^3+15ax^2-14ax<BR>\n" );
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document.write( "There is no easy way.  You find the combination that works.<BR>\n" );
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document.write( "In this one, all 3 terms have an a & and x.  Factor those out 1st.<BR>\n" );
document.write( "That leaves 9x^2 + 15x - 14<BR>\n" );
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document.write( "It'll be<BR>\n" );
document.write( "(ax + b)*(cx + d)<BR>\n" );
document.write( "a*b = 9, so that's either 1*9 or 3*3<BR>\n" );
document.write( "b*d = 14, so that's either 1*14 or 2*7 with different signs (one +, one minus)<BR>\n" );
document.write( "Then you find a combination that gives +15x<BR>\n" );
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