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

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