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.

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! :)
