Question 104461
{{{p(x)=60x(1+x)^72-(1+x)^72+1=0}}}
Let's move some terms around and discuss,
{{{p(x)=(1+x)^72*(60x-1)+1}}}
Since {{{(1+x)^72}}} is always positive or zero, let call that A(x) where A(x) is always positive except when x=-1 and A(-1)=0. 
{{{p(x)=A(x)*(60x-1)+1}}}
Let’s talk about this function for very large positive and negative x. 
For even modest values of x (x<-2 and x>0), A(x) grows rapidly because of the power of 72. 
For large positive x, A(x) gives a very large, positive number, (60x-1) gives a large, positive number, therefore p(x) is a very large, positive number since positive times a positive equals a positive. 
For large negative x, A(x) is a very large, positive number, (60x-1) is a large negative number, therefore p(x) is a very large, negative number since positive times a negative equals a negative. 
Therefore between large positive x and large negative x comes a point where p(x)=0, that is, it has a root.
Or using the vocabulary of the IVT, 
Since
{{{lim( x->infinity, p(x) ) 
= infinity}}}
{{{lim( x->-infinity, p(x) ) 
= -infinity}}}
There exists an x in the interval from {{{-infinity}}} to {{{infinity}}} where {{{p(x)=0}}}.