Real solutions to (x²+2x+3)(x²+x+1)(5x+3)=1001

[anemone]

Find all real solution(s) for the equation $(x^2+2x+3)(x^2+x+1)(5x+3)=1001$.

 

[RLBrown]

Spoiler

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

 

[anemone]

Thanks, RLBrown for participating in my challenge, but...

Spoiler

...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 $x$?

My solution:

Spoiler

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.