Suppose you are faced with the following problem:<\/p>\n
Solve -6x5<\/sup>+46x3<\/sup>+72x=0<\/p>\n This problem, when broken into the right steps, is not too hard. However, the steps are numerous.<\/p>\n Step 1:<\/p>\n Like most problems you encounter that contain x2<\/sup> or higher, you must factor. At the beginning of every factoring process comes the question: Is there a common factor?<\/em> (For a crash course in common factoring, look here<\/a>.)<\/p>\n We realize that, in our case, there indeed is a common factor: -2x. (As I mention in my post about common factoring<\/a>, whenever the lead term is negative, you should factor the negative out.)<\/p>\n After the common factor is taken out front, we arrive at<\/p>\n -6x5<\/sup>+46x3<\/sup>+72x=-2x(3x4<\/sup>-23x2<\/sup>-36)=0<\/p>\n Step 2:<\/p>\n We are now faced with how to factor the “inside”: 3x4<\/sup>-23x2<\/sup>-36. Of course, the common factor has already been removed. Since the lead coefficient is a 3, rather than a 1, we must use complex trinomial factoring<\/a>:<\/p>\n i) Examining 3x4<\/sup>-23x2<\/sup>-36, we multiply 3(-36) to get -108.<\/p>\n ii) We need to find two numbers that multiply to -108, but add to -23 (the middle term). We start writing down pairs of numbers that multiply to make 108.<\/p>\n Once your number pairs reverse to earlier combinations, you won’t get anything new.<\/p>\n iii) Among the numbers we’ve tabulated above, we must find the pair that, if one is negative, can also add to make -23 (the second term in 3x4<\/sup>-23x2<\/sup>-36). We notice the pair to be 4 and 27, if we make 27 negative. Once again: one of the pair has to be negative, since, as mentioned above, the pair must actually multiply to -108.<\/p>\n iv) We rewrite 3x4<\/sup>-23x2<\/sup>-36 with its middle term shown as the sum of 4 and -27:<\/p>\n 3x4<\/sup>-23x2<\/sup>-36=3x4<\/sup>+4x2<\/sup>-27x2<\/sup>-36<\/p>\n v) We separate the rewritten expression into two pairs, then common factor each pair:<\/p>\n (3x4<\/sup>+4x2<\/sup>)+(-27x2<\/sup>-36)=x2<\/sup>(3x2<\/sup>+4)-9(3x2<\/sup>+4)<\/p>\n vi) The repeating factor (3x2<\/sup>+4) indicates we are successful. Now, we reorganize our factored expression into two brackets:<\/p>\n x2<\/sup>(3x2<\/sup>+4)-9(3x2<\/sup>+4)=(x2<\/sup>-9)(3x2<\/sup>+4)<\/p>\n Step 3: We rewrite our original equation in factored form:<\/p>\n -6x5<\/sup>+46x3<\/sup>+72x=-2x(x2<\/sup>-9)(3x2<\/sup>+4)<\/p>\n We now notice the term (x2<\/sup>-9). Being a difference of squares<\/a>, it can be factored to (x+3)(x-3). Finally, we arrive at<\/p>\n -2x(x+3)(x-3)(3x2<\/sup>+4)=0<\/p>\n Step 4: We need to solve the equation: we need to report the values of x that will make the left side equal to 0. We use the following reasoning:<\/p>\n If several values multiply to make zero, one of them must already be zero.<\/em><\/p>\n Therefore,<\/p>\n -2x(x+3)(x-3)(3x2<\/sup>+4)=0<\/p>\n implies that either -2x=0, or (x+3)=0, or (x-3)=0, or (3x2<\/sup>+4)=0. One by one, we consider each possibility:<\/p>\n If -2x=0, then x=0. Therefore, x=0 is one solution.<\/p>\n If (x+3)=0, then x=-3. Therefore, x=-3 is a solution.<\/p>\n If (x-3)=0, then x=3. Therefore, x=3 is a solution.<\/p>\n Since, in the real numbers, x2<\/sup> cannot be negative, (3x2<\/sup>+4) cannot be equal to zero. It yields no solutions.<\/p>\n Our solutions to the seemingly endless problem -6x5<\/sup>+46x3<\/sup>+72x=0 are, finally, x=0, x=3, and x=-3.<\/p>\n You wouldn’t see many problems this difficult on a high school exam; you might encounter one. Good luck with it!<\/p>\n\n\n
\n 1<\/td>\n 108<\/td>\n<\/tr>\n \n 2<\/td>\n 54<\/td>\n<\/tr>\n \n 3<\/td>\n 36<\/td>\n<\/tr>\n \n 4<\/td>\n 27<\/td>\n<\/tr>\n \n 6<\/td>\n 18<\/td>\n<\/tr>\n \n 9<\/td>\n 12<\/td>\n<\/tr>\n \n 12<\/td>\n 9<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n