MHB Find Real Solutions for Equation $(x^2+2x+3)(x^2+x+1)(5x+3)=1001$

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Find all real solution(s) for the equation $(x^2+2x+3)(x^2+x+1)(5x+3)=1001$.
 
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This factors as
(-2 + x) (496 + 233 x + 95 x^2 + 28 x^3 + 5 x^4)
2 is the only real root, the second factor is strictly increasing
 
Thanks, RLBrown for participating in my challenge, but...
...do you mind to tell me more how we are going to tell, perhaps offhand, that $496 + 233 x + 95 x^2 + 28 x^3 + 5 x^4$ is strictly increasing over the real $x$?:confused:

My solution:
I first let $f(x)=((x^2+2x+3))((x^2+x+1)(5x+3))$ and I then find its first derivative

$\begin{align*}f'(x)&=((x^2+2x+3))(5(x^2+x+1)+(2x+1)(5x+3))+(2x+1)((x^2+x+1)(5x+3))\\&=25x^4+72x^3+117x^2+86x+30\\&=(25x^4+72x^3+55x^2)+(62x^2+86x+30)\\&=x^2\left(\left(x+\dfrac{36}{25}\right)^2+\dfrac{79}{625}\right)+62\left(\left(x+\dfrac{43}{62}\right)^2+\dfrac{11}{3844}\right)\\&>0\end{align*}$

and notice that $f'(x)$ is always greater than $0$ and hence $f$ is an increasing function.

We can conclude partially that the original equation $(x^2+2x+3)(x^2+x+1)(5x+3)=1001$ has only one real solution.

Through the prime factorization for $1001=7(11)(13)$, it is not hard to see that $5x+3=13\implies x=2$ is the answer.
 
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