document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=182397>Question 182397</A>: <I><SPAN STYLE=\"word-break: break-all;\">  Factor each polynomial. a^2-2a-35\r<BR>\n" );
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document.write( "Factor completely. -3t^3+ 3t^2-6t\r<BR>\n" );
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document.write( "Factor polynomial completely. 10a^2+ab-2b^2\r<BR>\n" );
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document.write( "Factor completely. 4m^2+20m+25 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #137002 by </B><A HREF=/tutors/aboutme.mpl?userid=J2R2R><B>J2R2R(94)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=J2R2R><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=J2R2R><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=137002\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Factor each polynomial. a^2-2a-35<BR>\n" );
document.write( "Here, the two number inside the brackets must add up to -2 and multiply together to give -35<BR>\n" );
document.write( "-7 and +5 satisfy the criteria<BR>\n" );
document.write( "a^2-2a-35 = (a-7)(a+5)\r<BR>\n" );
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document.write( "Factor completely. -3t^3+ 3t^2-6t <BR>\n" );
document.write( "Here we have a common factor -3t<BR>\n" );
document.write( "-3t^3+ 3t^2-6t = -3t(t^2-t+2)<BR>\n" );
document.write( "(t^2-t+2) cannot be factored any further (using real numbers).\r<BR>\n" );
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document.write( "Factor polynomial completely. 10a^2+ab-2b^2 <BR>\n" );
document.write( "Here we have the two coefficient numbers multiplied together to give 10 (from 10a^2) and the two other numbers multiplied together give -2 (from -2b^2)<BR>\n" );
document.write( "Then we have a fiddling act where crossover multiplying and adding together give 1 (from ab = 1ab).<BR>\n" );
document.write( "5a and 2a are useful to start with and 2 and 1 (with one of these being a minus which we will decide on the next line).<BR>\n" );
document.write( "You should see that 5*1 and 2*2 when a minus is in front of the 2 will make 5*1-2*2=1, so the 2 has a minus. These are the crossovers (5a with b and -2b with 2a)<BR>\n" );
document.write( "10a^2+ab-2b^2 = (5a-2b)(2a+b)\r<BR>\n" );
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document.write( "Factor completely. 4m^2+20m+25<BR>\n" );
document.write( "Similarly to above but just observe that 4m^2 and 25 are both squares, so think of using 2m and 5.<BR>\n" );
document.write( "2m times 2m gives 4m^2 and 5 times 5 gives 25. Now the crossovers 2m times 5 = 10m and the same again the other way 5 times 2m = 10m, combined makes 20m, so we have<BR>\n" );
document.write( "4m^2+20m+25=(2m+5)(2m+5) = (2m+5)^2<BR>\n" );
document.write( "Whenever there is a square factor, just double the crossover since it is the sum of the product of the two numbers in both directions e.g. 2 times 5 plus 5 times 2 or simply 2 times 2 times 5. With this in mind you should be able to spot square factors like this. 2^2=4, 5^2=25, 2x2x5=20.<BR>\n" );
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