efficient method to compare polynomials (fixed degree and non-negative indeterminate $x$) Let's say i have two polynomials $P_1(x)=a+bx+cx^2+dx^3$ and $P_2(x)=p+qx+rx^2+sx^3$
where $a,b,c,d \ge 0$ and $p,q,r,s \ge 0$
$a,b,c,d,p,q,r,s$ are all integers
then can i say
if $a+2b+4c+8d \gt p+2q+4r+8s$
then can I say $a+bt+ct^2+dt^3 \ge p+qt+rt^2+st^3$ for $t \ge 1$?
$t$ is an integer.
All I want to do is minimize the calculation part in my algo.
 A: No. Take $(a,b,c,d,p,q,r,s)=\left(0,\frac 32,0,2,0,1,2,1\right)$.
We have 
$$a+2b+4c+8d\gt p+2q+4r+8s$$
but 
$$a+bt+ct^2+dt^3\color{red}{\lt} p+qt+rt^2+st^3$$
for $t=1$.

Take $(a,b,c,d,p,q,r,s)=(1,1,2,1,0,0,0,2)$ where $a,b,c,d,p,q,r,s$ are non-negative integers.
We have
$$a+2b+4c+8d\gt p+2q+4r+8s$$
but 
$$a+bt+ct^2+dt^3\color{red}{\lt} p+qt+rt^2+st^3$$
for $t=3$.
A: Your question is equivalent to:

If $P$ is a cubic polynomial with integer coefficients such that $P(2) \ge0$, does it follow that $P(t)\ge0$ for all $t \in \mathbb N$?

This is clearly not true. Take $P(x)=-(x-2)^3$.
A: $$a+2b+4c+8d > p+2q+4r+8s$$
Let $t=1$,
$$a+b+c+d < p+q+r+s$$
Let $a=b=c=d=1$.
$$15 > p+2q+4r+8s$$
$$4 < p+q+r+s$$
Let me set $p=4, q=1, r=0, s=0$ and I have proven that your claim is not true.
A: Taking into account your comment about how you intended to use the
relationship that you had conjectured, your purpose was to find an
easily-tested relationship among the coefficients of the two polynomials
$P_1(x) = a+bx+cx^2+dx^3$ and $P_2(x) = p+qx+rx^2+sx^3$
along with a minimum parameter value $m$ such that if the 
coefficients satisfy your condition, then you can guarantee that
$$ a+bx+cx^2+dx^3 > p+qx+rx^2+sx^3 \tag1$$
for every integer $x$ such that $x \geq m.$
(You were attempting to use the minimum value $m=1,$ but since you are
mainly concerned about having to evaluate the polynomials for thousands
of values of $x$ I think we can afford to allow the value of $m$ to be
greater than $1$ as long as it is not too large.)
The desired inequality $(1)$ is equivalent to
$$ P_3(x) = (a-p) + (b-q)x + (c-r)x^2 + (d-s)x^3 > 0.$$
Now a fact about polynomials is that if a polynomial
(in this case $P_3(x)$) is not constant,
it tends either to $+\infty$ or to $-\infty$ as the parameter $x$
tends to $+\infty.$
In the first case there is some minimum value $m$ such that
$P_3(x) > 0$ whenever $x \geq m$;
in the second case there is some value $m$ such that
$P_3(x) < 0$ whenever $x \geq m.$
And you only need to look at leading term to see which way the 
polynomial will go.
For $P_3(x),$ as long as $d \neq s$ then the leading term is
$(d-s)x^3,$ and in order to have $P_3(x) > 0$ whenever $x \geq m$
we just need to see that $d > s.$
Once we see that, it remains only to find a suitable value of $m.$
When $d > s,$ to simplify things, let
\begin{align}
d' &= d - s, \\
p' &= \begin{cases} p - a & p > a, \\  0 & p \leq a, \end{cases}\\
q' &= \begin{cases} q - b & q > b, \\  0 & q \leq b, \end{cases}\\
r' &= \begin{cases} r - c & r > c, \\  0 & r \leq c, \end{cases}
\end{align}
and set $P_4(x) = -p' - q'x - r'x^2 + d'x^3.$
Then $a - p \geq -p' \geq 0,$ $b - q \geq -q' \geq 0,$ 
$c - r \geq r' \geq 0,$ and therefore
$P_3(x) \geq P_4(x)$ whenever $x \geq 0.$
Notice that whatever the value of $m$ turns out to be, if
$x \geq m$ then we can set $y = x - m,$ 
which implies that $y \geq 0.$ Then
\begin{align}
P_4(x) &= -p' - q'(m+y) - r'(m+y)^2 + d'(m+y)^3 \\
&= -p' - q'(m+y) - r'(m^2 + 2my + y^2) + d'(m^3 + 3m^2y + 3my^2 + y^3) \\
&= -p' - mq' - m^2r' + m^3d' + (-q' - 2mr' + 3m^2d')y + (-r' + 3md')y^2 + d'y^3\\
&\geq -p' - mq' - m^2r' + m^3d' + \frac3m(-p' - mq' - m^2r' + m^3d')y \\
&\qquad + \frac3{m^2}(-p' - mq' - m^2r' + m^3d')y^2 + \frac1{m^3}(-p' - mq' - m^2r' + m^3d')y^3\\
&= \left(1 + \frac3my + \frac3{m^2}y^2 + \frac1{m^3}y^3\right)
(-p' - mq' - m^2r' + m^3d'),
\end{align}
and the last quantity is positive as long as $m^3d' - m^2r' - mq' - p' > 0.$
In summary, if we set $d',$ $p',$ $q',$ and $r'$ as shown above, and if
$m^3d' - m^2r' - mq' - p' > 0,$ then whenever $x\geq m$ it follows that
$P_4(x) >0,$ therefore $P_3(x) > 0,$
and therefore $a+bx+cx^2+dx^3 > p+qx+rx^2+sx^3.$
