Solve $x^{11}+x^8+5\equiv 0\pmod{49}$ Solve $x^{11}+x^8+5\pmod{49}$

My work
$f(x)=x^{11}+x^8+5$
consider the polynomial congruence $f(x) \equiv 0 \pmod {49}$
Prime factorization of $49 = 7^2$
we have $f(x) \equiv 0 \mod 7^2$
Test the value $x\equiv0,1,2,3,4,5,6$ for $x^{11}+x^8+5 \equiv 0\pmod 7$
It works for $x\equiv1$, $x\equiv1\pmod7$ is a solution.
We proceed to lift the solution $\mod 7^2$
$f'(x) = 11x^{10} + 8x^7 +5$, we have $f(1) = 7$ , $f'(x) = 24$
Since $7$ can not divide $f'(1)$, we need to solve $24t \equiv 0 \mod 7$
we get $t \equiv 0 \pmod 5$
and then ..................... ?????
 A: Hint $\rm\ mod\ 49\!:\ 5+(1\!+\!7n)^8\!+(1\!+\!7n)^{11}\!\equiv 5 + (1\!+\!56n) + (1\!+\!77n) \equiv 7 - 14 n\equiv 7(1\!-\!2n)$
Thus $\rm\ 49\mid 7\,(1\!-\!2n)\iff 7\mid 1\!-\!2n\iff n\equiv 4\,\ (mod\ 7),\ $ so $\rm\ x \equiv 1+7n\equiv 29\,\ (mod\ 49).$
Alternatively $ $ we may compute $\rm\:f(1\!+\!7n)\:$ by Taylor's formula (vs. Binomial Theorem above)
$$\rm mod\ 49\!:\,\ g(n) = f(1\!+\!7n) = g(0) + g'(0)\, n + \cdots\, \equiv\, f(1) + 7\, f'(1)\,n$$
Thus $\rm\ 7\mid f(1)\:\Rightarrow\:49\mid f(1)+7\,f'(1)\,n\!\iff\! 7\mid f(1)/7+f'(1)\,n\!\iff\! n\equiv\, -\dfrac{f(1)/7}{f'(1)}\:\ (mod\ 7)$
So $\rm\,\ mod\ 7\!:\ n \equiv -\dfrac{f(1)/7}{f'(1)} \equiv \dfrac{-1}{11+8}\equiv \dfrac{6}{-2}\equiv -3\equiv 4.\:$ This is equivalent to using Hensel's Lemma.  
It is instructive to compare the two approaches.
A: If $x = 1 + 7 t$ for an integer $t$, then $x^{11} \equiv 1 + 11 (7 t) \mod 49$, 
$x^8 \equiv 1 + 8 (7 t) \mod 49$, and so $f(x) \equiv 7 + 19 (7t) \mod 49$.
Thus you want $1 + 19 t \equiv 0 \mod 7$, ...
A: Why not use Hensel's Lemma ?
First, let us find a solution modulo $\,7\,$ of $\,f(x)=x^{11}+x^8+5\pmod 7\,$.
Clearly, $\,r=1\,$ is a root. Now, define
$$t:=-\frac{f(1)}{7}f'(1)^{-1}=-(11+8)^{-1}=(-19)^{-1}=30^{-1}=18\pmod{49}\Longrightarrow $$
$$s:=1+18\cdot 7=127=29\pmod{49}\,\,\,\text{is a root of}\,\,\,f(x)\pmod{49}$$
What is cool in this method is that it allows you find roots modulo $\,7^2\,,\,7^3\,\ldots\,,\,7^k\,$ , for any $\,k\in\Bbb N\,$ , as long as you have one simple root modulo $\,7\,$ (why does it have to be "simple"?)
