document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=723322>Question 723322</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi what are the zeros of these functions?                                                                            F(x)=3x^2+x-2                                                                                                                                                F(x)=x^3+6x^2+16                                                                                                                                  F(x)=x^3+8x^2-20x                                                                                                                                      F(x)=x^3-7x^2+7x+15 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #443167 by </B><A HREF=/tutors/aboutme.mpl?userid=DrBeeee><B>DrBeeee(684)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=DrBeeee><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=443167\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Your first function is<BR>\n" );
document.write( "(1) 3x^2 + x - 2 = 0, factors into<BR>\n" );
document.write( "(2) (3x - 2)(x + 1) = 0, set the factors equal to zero and solve for the root x give<BR>\n" );
document.write( "(3) 3x - 2 = 0 or<BR>\n" );
document.write( "(4) x = 2/3 and<BR>\n" );
document.write( "(5) x + 1 = 0 gives us<BR>\n" );
document.write( "(6) x = -1 <BR>\n" );
document.write( "Let's check the roots using (1).<BR>\n" );
document.write( "Is (3*(2/3)^2 + 2/3 -2 =0)?<BR>\n" );
document.write( "Is (4/3 + 2/3 -2 = 0)?<BR>\n" );
document.write( "Is (6/3 - 2 = 0)?<BR>\n" );
document.write( "Is (2 - 2 = 0)? Yes<BR>\n" );
document.write( "Is (3(-1)^2 - 1 - 2 = 0)?<BR>\n" );
document.write( "Is (3 - 3 = 0)? Yes<BR>\n" );
document.write( "The roots of (1) are {-1,2/3}<BR>\n" );
document.write( "Your second function is<BR>\n" );
document.write( "(7) x^3 + 6x^2 + 16 = 0<BR>\n" );
document.write( "Because (7) is an odd order polynomial, it must have at least one real root. You can find it by trial and check. I found the real root of (7) to approximately<BR>\n" );
document.write( "(8) x = -6.391648<BR>\n" );
document.write( "I don't know if this can be derived analytically or not. To find the remaining quadratic, divide (7) by (x+6.39...) and get<BR>\n" );
document.write( "(9) (x^2 - (.39...)x + 2.5032...)<BR>\n" );
document.write( "The quadratic (9) has a complex pair of roots at approximately<BR>\n" );
document.write( "(10) x = 0.2 +or- 1.59i<BR>\n" );
document.write( "You'll have to check them - way too tired tonight. I know the real root is correct. If you have a graphing calculator, graph (7), it's really neat. It's derivatives are at x = zero and x = -4. The graph show how the curve of (7) changes \"direction\", slope equals zero, at -4 and 0.<BR>\n" );
document.write( "Your third function is<BR>\n" );
document.write( "(10) x^3 + 8x^2 - 20x = 0 which factors into<BR>\n" );
document.write( "(11) x(x + 10)(x - 2) = 0 giving three real roots<BR>\n" );
document.write( "(12) x = {0,-10,2}<BR>\n" );
document.write( "You can easily show that each value of in (12) satisfies (10).<BR>\n" );
document.write( "Your fourth function is<BR>\n" );
document.write( "(13) x^3 - 7x^2 + 7x + 15 = 0<BR>\n" );
document.write( "It also must have one real root so let's find it. I start by letting the sum of last two terms equal to zero or<BR>\n" );
document.write( "(14) 7x + 15 = 0 or<BR>\n" );
document.write( "(15) x = -15/7 or x is about -2. Let's try x = -2.<BR>\n" );
document.write( "Is ((-2)^3 - 7(-2)^2 + 7(-2) + 15 = 0)?<BR>\n" );
document.write( "Is (-8 - 28 -14 +15 = 0)?<BR>\n" );
document.write( "Is (-35 = o)? not even close! But we know that the rel root is between -2 and 0, because f(-2)<0 and f(0)=15>0. So let's try x = -1<BR>\n" );
document.write( "Is ((-1)^3 - 7(-1)^2 + 7(-1) + 15 = 0)?<BR>\n" );
document.write( "Is (-1 -7 -7 +15 = 0)? Yes, we found the real root!!<BR>\n" );
document.write( "(16) x = -1<BR>\n" );
document.write( "Divide (13) by (x+1) to find the quadratic, and get<BR>\n" );
document.write( "(17) x^2 -8x +15 = 0 which factors to<BR>\n" );
document.write( "(18) (x-5)(x-3) = 0 and the roots are<BR>\n" );
document.write( "(19) x = {5,3}, and for your fourth function (13) the roots are<BR>\n" );
document.write( "(20) x = (-1,3,5}<BR>\n" );
document.write( "PS My roots are your zeroes.\r<BR>\n" );
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