Section+2.4

(ANswers provided by Matthew, Jenn, and Dimitri) || Student Feedback || Teacher Feedback || f(2)=7 is the point (2,7) which does not justify a real root ... the graph justifies a real root at the x value of -1.116 ||
 * Student Answers
 * 63. || False, if f(x) had any factor (x+2) then f(2) would still be a positive answer (assuming the other factors are as well positive). Regardless for f(anything) to = 0, the only answer likely for the given situation would be -2, not 2. || correct ||
 * 64. || True. When using long division to divide a problem, the resulting factors will be multiplied (hence the term... factors) and any remaining terms (remainders) are added. 3, a remainder, is added to the factors, (x-1) and (2x^2-x+1) || correct ||
 * 65. || A (x-3 would equal 0, not x+3) || Correct ||
 * 66. || E, using a calculator and the "calculate --> zero" program, all roots with the exception of E are valid || Correct ||
 * 67. || B, the remainder is -3 when f(x) is divided by (x+2) not (x-2) || Correct ||
 * 68 || E, by proving C as possible, we prove that real root(s) exist! || Answer is correct BUT explanation is incorrect.