document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209697>Question 1209697</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let f(x) = x^3 + ax^2 + bx + c be a polynomial.  All the roots are negative integers.  If a + b + c = 10, then find c. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849828 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=849828\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the roots of $f(x)$ be $-r_1, -r_2, -r_3$, where $r_1, r_2, r_3$ are positive integers.<BR>\n" );
document.write( "Then $f(x) = (x+r_1)(x+r_2)(x+r_3) = x^3 + (r_1+r_2+r_3)x^2 + (r_1r_2 + r_1r_3 + r_2r_3)x + r_1r_2r_3$.<BR>\n" );
document.write( "Comparing this with $f(x) = x^3 + ax^2 + bx + c$, we have:<BR>\n" );
document.write( "$a = r_1+r_2+r_3$<BR>\n" );
document.write( "$b = r_1r_2 + r_1r_3 + r_2r_3$<BR>\n" );
document.write( "$c = r_1r_2r_3$\r<BR>\n" );
document.write( "\n" );
document.write( "We are given $a+b+c = 10$, so<BR>\n" );
document.write( "$r_1+r_2+r_3 + r_1r_2 + r_1r_3 + r_2r_3 + r_1r_2r_3 = 10$.<BR>\n" );
document.write( "Adding 1 to both sides, we get<BR>\n" );
document.write( "$1 + r_1+r_2+r_3 + r_1r_2 + r_1r_3 + r_2r_3 + r_1r_2r_3 = 11$<BR>\n" );
document.write( "$(1+r_1)(1+r_2)(1+r_3) = 11$<BR>\n" );
document.write( "Since $r_1, r_2, r_3$ are positive integers, $1+r_1, 1+r_2, 1+r_3$ must be integers greater than 1.<BR>\n" );
document.write( "Since 11 is a prime number, the only possible factorization is $11 = 11 \cdot 1 \cdot 1$.<BR>\n" );
document.write( "Thus, we must have $1+r_1 = 11$, $1+r_2 = 1$, and $1+r_3 = 1$. This means $r_1 = 10$, $r_2 = 0$, and $r_3 = 0$. However, the roots must be positive integers, so $r_2$ and $r_3$ cannot be 0.\r<BR>\n" );
document.write( "\n" );
document.write( "Alternatively, we can have $1+r_1 = 11$, which means $r_1 = 10$. Then $(1+r_2)(1+r_3) = 1$, which is not possible since $r_2$ and $r_3$ are positive integers.\r<BR>\n" );
document.write( "\n" );
document.write( "If we let $1+r_1=11$, $1+r_2=1$ and $1+r_3=1$, then $r_1=10$, $r_2=0$ and $r_3=0$, which are not all positive integers.<BR>\n" );
document.write( "However, if we consider the case where two of the integers are 1, then we have<BR>\n" );
document.write( "$(1+r_1)(1+r_2)(1+r_3) = 11$.<BR>\n" );
document.write( "Let $r_2 = r_3 = 0$. Then $1+r_1 = 11$, so $r_1 = 10$.<BR>\n" );
document.write( "$a = 10+0+0 = 10$<BR>\n" );
document.write( "$b = 0+0+0 = 0$<BR>\n" );
document.write( "$c = 0$<BR>\n" );
document.write( "$a+b+c = 10+0+0 = 10$.<BR>\n" );
document.write( "So $c = 0$.\r<BR>\n" );
document.write( "\n" );
document.write( "Final Answer: The final answer is $\boxed{0}$<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );