Expert: Sherry Wallin - 10/8/2010

Question
6x^3+mx^2+nx-5 has a factor of x+1. When divided by x-1, it
has remainder of -4. What are the values of m and n?

Answer
Mike~

First let me tell you I never received an email with this question. I was out on the website at allexperts.com answering someone else's question and saw yours here.

There is a lot of information in this problem. But first do the long division and see that you have a remainder of -4 in this way:


       6x^2 + (m+6)x + (n+m+6)
      ___________________________
x - 1 | 6x^3 + mx^2 + nx - 5
     -(6x^3 - 6x^2)
       -------------
         (m+6)x^2 + nx
       -[(m+6)x^2 -(m+6)x]
       ---------------------
                  [n+(m+6)x] - 5
                -[(n+m+6)x -(n+m+6)]
                -----------------
                           n+m+6-5 = n+m + 1 => this is your remainder

which by the way, is suppose to equal -4:

n + m + 1 = -4 => n + m = -5

Now that you know that the original expression has a factor of x + 1 divide it  6x^3 + mx^2 + nx - 5, by x + 1 and set that remainder equal to 0 and then you will have two equations in m and n and you can solve the system.

Please let me know if you can't do this or want me to check your answer. Good Luck.

Math Prof