Proof that given equation(quartic) doesn't have real roots $$
(x^2-9)(x-2)(x+4)+(x^2-36)(x-4)(x+8)+153=0
$$
I need to prove that the above equation doesn't have a real solution. I tried breaking it up into an $(\alpha)(\beta)\cdots=0$ expression, but no luck. Wolfram alpha tells me that the equation doesn't have real roots, but I'm sure there's simpler way to solve this than working trough the quartic this gives.
 A: Your polynomial is
$$P(x) = ({x^2} - 9)(x - 2)(x + 4) + ({x^2} - 36)(x - 4)(x + 8) + 153\tag{1}$$
Now consider theses
$$\eqalign{
  & f(x) = ({x^2} - 9)(x - 2)(x + 4)  \cr 
  & f({x \over 2}) = \left( {{{\left( {{x \over 2}} \right)}^2} - 9)} \right)\left( {\left( {{x \over 2}} \right) - 2} \right)\left( {\left( {{x \over 2}} \right) + 4} \right)  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over 4}\left( {{x^2} - 36} \right){1 \over 2}\left( {x - 2} \right){1 \over 2}\left( {x - 8} \right)  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {16}}\left( {{x^2} - 36} \right)\left( {x - 2} \right)\left( {x - 8} \right) \cr}\tag{2}$$
combine $(1)$ and $(2)$ to get
$$P(x) = f(x) + 16f({x \over 2}) + 153\tag{3}$$
Next, we try to find the range of $f(x)$. For this purpose, consider this
$$\eqalign{
  & f(x) = ({x^2} - 9)(x - 2)(x + 4)  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\, = {x^4} + 2{x^3} - 17{x^2} - 18x + 72  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\, = {\left( {{x^2} + x - 9} \right)^2} - 9 \cr}\tag{4}$$
Now, by $(4)$ you can conclude that
$$\left\{ \matrix{
  f(x) \ge  - 9 \hfill \cr 
  f({x \over 2}) >  - 9 \hfill \cr}  \right.\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\left\{ \matrix{
  f(x) >  - 9 \hfill \cr 
  f({x \over 2}) \ge  - 9 \hfill \cr}  \right.\tag{5}$$
Notice the equality signs! Can you figure out why this happens? Then using $(5)$ you can conclude that
$$\left\{ \matrix{
  f(x) \ge  - 9 \hfill \cr 
  16f({x \over 2}) >  - 144 \hfill \cr}  \right.\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\left\{ \matrix{
  f(x) >  - 9 \hfill \cr 
  16f({x \over 2}) \ge  - 144 \hfill \cr}  \right.\tag{6}$$
and then summing up either of the relations $(6)$ will lead to
$$f(x) + 16f({x \over 2}) >  - 153\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,f(x) + 16f({x \over 2}) + 153 > 0\,\,\,\,\,\, \to \,\,\,\,\,\,\,P(x) > 0\tag{7}$$
I think we are done now! :)
A: Expand $x^2-9=(x-3)(x+3)$ and $x^2-36=(x-6)(x+6)$. Let $t=x+\frac 1 2$, $u=x+1$. Then
$$LHS=(t^2-(2.5)^2)(t^2-(3.5)^2)+(u^2-25)(u^2-49)+153\geq -(\frac {3.5^2-2.5^2} 2)^2-(\frac {49-25} 2)^2+153=0$$
with equality iff $t^2=\frac {2.5^2+3.5^2} 2$ and $u^2=\frac {25+49} 2$ which is incompatible. Hence $LHS>0$
