document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209706>Question 1209706</A>: <I><SPAN STYLE=\"word-break: break-all;\"> There are integers b, c for which both roots of the polynomial x^2  - x - 3 are also roots of the polynomial x^3 - bx^2 - c. Determine the ordered pair (b,c).<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 #849859 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1987)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=849859\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $P(x) = x^2 - x - 3$. The roots of $P(x)$ are given by the quadratic formula:<BR>\n" );
document.write( "$$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-3)}}{2(1)} = \frac{1 \pm \sqrt{1+12}}{2} = \frac{1 \pm \sqrt{13}}{2}$$<BR>\n" );
document.write( "Let $r_1 = \frac{1 + \sqrt{13}}{2}$ and $r_2 = \frac{1 - \sqrt{13}}{2}$.<BR>\n" );
document.write( "Since $r_1$ and $r_2$ are roots of $x^2 - x - 3 = 0$, we have<BR>\n" );
document.write( "$r_1^2 - r_1 - 3 = 0$ and $r_2^2 - r_2 - 3 = 0$.<BR>\n" );
document.write( "Thus, $r_1^2 = r_1 + 3$ and $r_2^2 = r_2 + 3$.<BR>\n" );
document.write( "Also, $r_1 + r_2 = 1$ and $r_1 r_2 = -3$.\r<BR>\n" );
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document.write( "Let $Q(x) = x^3 - bx^2 - c$. Since $r_1$ and $r_2$ are roots of $Q(x)$, we have<BR>\n" );
document.write( "$r_1^3 - br_1^2 - c = 0$ and $r_2^3 - br_2^2 - c = 0$.<BR>\n" );
document.write( "$r_1^3 = r_1 \cdot r_1^2 = r_1(r_1+3) = r_1^2 + 3r_1 = (r_1+3) + 3r_1 = 4r_1 + 3$<BR>\n" );
document.write( "$r_2^3 = r_2 \cdot r_2^2 = r_2(r_2+3) = r_2^2 + 3r_2 = (r_2+3) + 3r_2 = 4r_2 + 3$<BR>\n" );
document.write( "Substituting these into the equations for $Q(x)$, we get<BR>\n" );
document.write( "$4r_1 + 3 - b(r_1 + 3) - c = 0$<BR>\n" );
document.write( "$4r_2 + 3 - b(r_2 + 3) - c = 0$<BR>\n" );
document.write( "Subtracting the second equation from the first, we get<BR>\n" );
document.write( "$4(r_1 - r_2) - b(r_1 - r_2) = 0$<BR>\n" );
document.write( "$r_1 - r_2 \ne 0$, so $4 - b = 0$, which gives $b = 4$.<BR>\n" );
document.write( "Substituting $b=4$ into the first equation, we get<BR>\n" );
document.write( "$4r_1 + 3 - 4(r_1 + 3) - c = 0$<BR>\n" );
document.write( "$4r_1 + 3 - 4r_1 - 12 - c = 0$<BR>\n" );
document.write( "$-9 - c = 0$<BR>\n" );
document.write( "$c = -9$<BR>\n" );
document.write( "Thus, $(b, c) = (4, -9)$.\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{(4,-9)}$<BR>\n" );
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