document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1207338>Question 1207338</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Show that x^2 + 4 is prime.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #846375 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=846375\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Let's look at an example or two before diving into this current question.\r<BR>\n" );
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document.write( "Example 1: Find the factorization of x^2+5x+6 over the real numbers.<BR>\n" );
document.write( "Solution: (x+3)(x+2)<BR>\n" );
document.write( "We can use trial-and-error to find the factors. We're looking for two numbers that add to 5 and multiply to 6.<BR>\n" );
document.write( "This is because (x+p)(x+q) = x^2+(p+q)x+p*q has the x coefficient p+q while the constant term is p*q.<BR>\n" );
document.write( "Since x^2+5x+6 factors to (x+3)(x+2), it is not prime over the real numbers.<BR>\n" );
document.write( "A prime polynomial only has factors of 1 and itself. \r<BR>\n" );
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document.write( "Example 2: is x^2+7x+10 prime over the set of real numbers? If no, then what is the factorization?<BR>\n" );
document.write( "Solution: No it is not prime. It factors to (x+5)(x+2)<BR>\n" );
document.write( "Again we could use trial-and-error to find the factors.<BR>\n" );
document.write( "Note how 2+5 = 7 and 2*5 = 10.<BR>\n" );
document.write( "Another approach that is more methodical without guess-and-check is to use the quadratic formula.<BR>\n" );
document.write( "Use that formula to solve x^2+7x+10 = 0 for x. I'll skip steps and leave them for the student to do.<BR>\n" );
document.write( "The two solutions are x = -5 and x = -2.<BR>\n" );
document.write( "x = -5 leads to x+5 = 0 which gives the factor x+5 <BR>\n" );
document.write( "x = -2 leads to x+2 = 0 which gives the factor x+2 \r<BR>\n" );
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document.write( "Example 3: Is x^2-7 prime over the rational numbers? If no, then what is the factorization?<BR>\n" );
document.write( "Solution: Yes it is prime over the rational numbers.<BR>\n" );
document.write( "Solve x^2-7 = 0 to get x = sqrt(7) and x = -sqrt(7)<BR>\n" );
document.write( "We arrive at factors ( x-sqrt(7) ) and ( x+sqrt(7) ), but sqrt(7) isn't in the set of rational numbers.<BR>\n" );
document.write( "Therefore we conclude that x^2-7 is prime over the rational numbers. However, it is not prime over the real numbers.\r<BR>\n" );
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document.write( "Now onto your current question.<BR>\n" );
document.write( "We want to show that x^2+4 is prime.<BR>\n" );
document.write( "This portion \"is prime\" is a bit vague since we haven't defined what set of numbers we want to work with. <BR>\n" );
document.write( "I'll assume your textbook is implying \"is prime over the real numbers\".\r<BR>\n" );
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document.write( "Solving x^2+4 = 0 leads to x = 2i and x = -2i where i = sqrt(-1) is an imaginary number.<BR>\n" );
document.write( "This means x-2i and x+2i are the factors.<BR>\n" );
document.write( "But these factors are not purely real numbers.\r<BR>\n" );
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document.write( "If your teacher asked you to factor x^2+4 over the complex numbers, then x^2+4 = (x-2i)(x+2i), which shows x^2+4 is not prime over the complex numbers.\r<BR>\n" );
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document.write( "But it is prime over the real numbers since the roots to x^2+4 = 0 are not in the set of real numbers.<BR>\n" );
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