How do you create a polynomial p which has degree 4, as #x->oo, p(x)->-oo#, p has exactly three x intercepts (-6,0), (1,0) and (117,0), the graph of #y=p(x)# crosses through the x axis at (1,0)?

1 Answer
Apr 2, 2017

Answer:

There are basically two possibilities:

#{ (p(x) = k(x+6)^2(x-1)(x-117)), (p(x) = k(x+6)(x-1)(x-117)^2) :}#

with #k < 0#

Explanation:

The leading coefficient of #p(x)# must be negative in order that #p(x)->-oo# as #x->oo#.

It must have linear factors #(x+6)#, #(x-1)# and #(x-117)#

As there are only three #x# intercepts, one of these must be repeated.

Also since #p(x)# crosses the #x# axis at #(1, 0)#, the corresponding factor #(x-1)# only occurs an odd number of times, and thus exactly once.

So there are basically two possibilities:

#{ (p(x) = k(x+6)^2(x-1)(x-117)), (p(x) = k(x+6)(x-1)(x-117)^2) :}#

with #k < 0#