document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=90546>Question 90546</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Factor each polynomial completely. If it cannot be factored write \"prime\"\r<BR>\n" );
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document.write( "              8x^2+28x-60\r<BR>\n" );
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document.write( "I have tried  4(2x^2+8x-15)<BR>\n" );
document.write( "              4(x+3)(x-5)<BR>\n" );
document.write( "  What do I do now!!  <BR>\n" );
document.write( " I can't get my -60 to come out when I check it.<BR>\n" );
document.write( "Please help!!!!!!!!  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #65724 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><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=65724\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Factor 8x^2+28x-60<BR>\n" );
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document.write( "Your first step was to take out a factor of 4. This is the correct thing to do. After that<BR>\n" );
document.write( "you are left with:<BR>\n" );
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document.write( "4*(2x^2 + 7x - 15)<BR>\n" );
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document.write( "The 2x^2 can only factor to 2x * x so the next step (if 2x^2 + 7x -15 factors) involves:<BR>\n" );
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document.write( "4*(2x ....)*(x ....)<BR>\n" );
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document.write( "the 15 can only factor to 15*1 or 5*3 (ignoring the signs on the two numbers). The potential<BR>\n" );
document.write( "factors of 2x^2 + 7x - 15 are therefore:<BR>\n" );
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document.write( "(2x 15)*(x 1)<BR>\n" );
document.write( "(2x 1)*(x 15)<BR>\n" );
document.write( "(2x 5)*(x 3)<BR>\n" );
document.write( "(2x 3)*(x 5)<BR>\n" );
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document.write( "If it factors, there must be some combination with signs between the x terms and the numbers<BR>\n" );
document.write( "that adds to +7. Becomes pretty apparent that the factors involving 15 are not likely.<BR>\n" );
document.write( "So the most likely factors are (2x 5)*(x 3) or (2x 3)*(x 5).<BR>\n" );
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document.write( "We could have:<BR>\n" );
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document.write( "(2x -5)*(x + 3) but this doesn't work because 2x*3 - 5x = 6x - 5x = +x not +7x<BR>\n" );
document.write( "or<BR>\n" );
document.write( "(2x + 5)*(x - 3) but this doesn't work because 2x*(-3) + 5x = -6x + 5x = -x not +7x<BR>\n" );
document.write( "so if it factors it must be either (2x + 3)*(x - 5) or (2x - 3)*(x + 5). Try:<BR>\n" );
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document.write( "(2x + 3)*(x - 5). The cross products are 2x*(-5) and 3*x = -10x + 3x = -7x. Close, but<BR>\n" );
document.write( "the sign is wrong.<BR>\n" );
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document.write( "We are down to trying (2x - 3)*(x + 5). The cross products are 2x*5 and -3*x which lead<BR>\n" );
document.write( "to 10x - 3x = +7x. That's what we are looking for.  So multiplying out the factors:<BR>\n" );
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document.write( "(2x - 3)*(x + 5) gives 2x*x + 2x*5 -3*x -3*5 = 2x^2 + 6x - 5x -15 = 2x^2 +7x - 15.<BR>\n" );
document.write( "Hooray! Now we can say that 2x^2 + 7x - 15 factors to (2x  - 3)*(x + 5) and as a result the<BR>\n" );
document.write( "whole factoring process results in:<BR>\n" );
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document.write( "4*(2x - 3)*(x + 5)<BR>\n" );
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document.write( "That's the answer and if you multiply this out the product will be the original polynomial <BR>\n" );
document.write( "you were given to factor. <BR>\n" );
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document.write( "Factoring like this involves a lot of judgment, guessing, and trial-and-error as you can<BR>\n" );
document.write( "see from the above analysis. The more you work these kind of problems the better you will <BR>\n" );
document.write( "become at getting to the answer quicker because you will develop a \"feel\" for what <BR>\n" );
document.write( "combinations  are more likely to work and what combinations are likely to lead to a <BR>\n" );
document.write( "wrong answer. <BR>\n" );
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document.write( "Hope you can follow the above process ... especially with regard to using the cross<BR>\n" );
document.write( "products (multiplying a term containing x by a term that is just a number) to look for<BR>\n" );
document.write( "a factor that works.<BR>\n" );
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document.write( "And hope that the answer above points out where you made a few simple mistakes in your <BR>\n" );
document.write( "logic that led you to an answer that wouldn't check.<BR>\n" );
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