document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=37023>Question 37023</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Factorization of c^3-64.\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 #22750 by </B><A HREF=/tutors/aboutme.mpl?userid=jcmtnez><B>jcmtnez(53)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jcmtnez><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=22750\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You may see that 8^2=64 and therefore (4*2)^2=64, that means that 4^2*2^2=64 (As probably now 2^2=4) therefore 4^2*4=64 that is the same that 4^3=64.<BR>\n" );
document.write( "As I show you 4 is a root of the polynomial c^3-64, and therefore when we divide by c-4 we will obtain another polynomial, at this time of degree two.\r<BR>\n" );
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document.write( "Using syntethetic division we easily have that (c^3-64)/(c-4)= c^2+4c+16, then we can express c^3-64 as (c-4)(c^2+4c+16) which is the factorized form.\r<BR>\n" );
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document.write( "Note: We can make more factorization because the solutions and the factors of c^2+4c+16 involve complex numbers. If you need more help, or you didn't understand something, please ask me again. </SPAN>\n" );
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