document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=211437>Question 211437</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need help factoring some equations please help:\r<BR>\n" );
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document.write( "2.	16xy + 32yz – 20xyz<BR>\n" );
document.write( "3.	9x2 – 64<BR>\n" );
document.write( "4.	5x3 – 125u2x <BR>\n" );
document.write( "5.	x2 + 7x + 12 \r<BR>\n" );
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document.write( "Please explain to me how to do these because I am confused and old. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #159812 by </B><A HREF=/tutors/aboutme.mpl?userid=drj><B>drj(1380)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=drj><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=drj><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=159812\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I need help factoring some equations please help:<BR>\n" );
document.write( "2. 16xy + 32yz – 20xyz<BR>\n" );
document.write( "3. 9x2 – 64<BR>\n" );
document.write( "4. 5x3 – 125u2x<BR>\n" );
document.write( "5. x2 + 7x + 12\r<BR>\n" );
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document.write( "Please explain to me how to do these because I am confused and old.\r<BR>\n" );
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document.write( "2.  16xy+32yz-20xyz.  Take a look at all three terms and find what is common.  That is, find the common factor.    \r<BR>\n" );
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document.write( "2a.  In this case, let's start with the common factor for the numbers 16, 32, -20.  The common factor is 4. How did I calculate this?  Find the lowest common factors: 16=2*2*2*2, 32=2*2*2*2*2, -20=-2*2*5.   The common factor is 2*2=4.  So factor out 4 in all three terms which yields:  4(4xy+8yz-5xyz)\r<BR>\n" );
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document.write( "2b.  Now, check if x is common in all three terms.  x is not a common factor.  So it cannot be factored out.\r<BR>\n" );
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document.write( "2c.  Now check if y is common to all three terms.  y is a common factor.  So factor out y.  4y(4x+8z-5xz). <BR>\n" );
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document.write( "2d.  Now check if z is common to all three terms.  z is not a common factor.\r<BR>\n" );
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document.write( "2e.  Therefore, final answer is 4y(4x+8z-5xz).\r<BR>\n" );
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document.write( "3. 9x2 – 64.  \r<BR>\n" );
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document.write( "3a.  This is a special case of the FOIL method.  That is, multiply (Ax-B)(Ax+B) where A=3 or Ax=3x and B=8.  FOIL means F=First O=Outer, I=Inner, L=Last:  Multiply First Terms (Ax*Ax), then Outer Terms, Ax*B, then Inner Terms, -B*Ax, then Last Terms, -B*B.  Therefore, \r<BR>\n" );
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document.write( "3b.  (Ax-B)(Ax+B)=Ax*Ax+Ax*B-B*Ax-B*B.  We can simplify this since the INNER and OUTER terms add up to zero.  Simplifying, yields (Ax-B)(Ax+B)=(Ax)^2-B^2. <BR>\n" );
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document.write( "3c.  So 9x2-64=(3x-8)(3x+8) is the final answer where A=3 or Ax=3x and B=8.\r<BR>\n" );
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document.write( "4. 5x3 – 125u2x.  Follows a similar process as Problem 2.\r<BR>\n" );
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document.write( "4a.  5x3-125u2x= 5(x3-25u2x).  Factored common factor 5.\r<BR>\n" );
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document.write( "4b.  5(x3-25u2x)=5x(x2-25u2).  Factors common factor x.  5x(x2-25u2) can be rewritten as 5x(x^2-25u^2), where u2 is meant as u-squared or u^2.\r<BR>\n" );
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document.write( "5. x2 + 7x + 12  x2+7x+12.  \r<BR>\n" );
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document.write( "5a.  Based on the FOIL method again, see above Problem 3.  You need to satisfy two equations:  A+B=7 and A*B=12.  So you need to find factors of 12 that add to seven.  You can use a trial and error method for this case.  \r<BR>\n" );
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document.write( "5b.  2 and 6 are factors of twelve but they don't add to seven.<BR>\n" );
document.write( "1 and 12 are factors of twelve but they don't add to seven.\r<BR>\n" );
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document.write( "5c.  Now, let's try 3 and 4.  When you multiply these numbers, you get 3*4=12.  When you add these numbers, you get 3+4=7.\r<BR>\n" );
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document.write( "5d.  Therefore, x2+7x+12=(x+3)(x+4) is the answer.  Now, use FOIL method to check answer:\r<BR>\n" );
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document.write( "(x+3)(x+4)=x*x+4x+3x+3*4=x2+7x+12.  So it works.\r<BR>\n" );
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document.write( "Note:  If you prefer a visual approach to factoring, I have developed free step-by-step videos (1-2 hours) on Factoring at http://www.FreedomUniversity.TV/courses/IntroAlgebra/Module5.html.  The set of videos have many similar problems as the ones that are describe above.\r<BR>\n" );
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