How to solve $ x^2+4x+2 \equiv 0 \pmod{49}$

Question

How to decompose equations below and then solve

1)$$ 2x^3 + 7x - 4 \equiv 0 \pmod{25} $$

2)$$ x^2+4x+2 \equiv 0 \pmod{49}$$

Thank you.

Answer 1

Completing the square $\rm\ mod\ 49\!:\ x^2\!+4x+2 = (x+2)^2\!-2 \equiv (x+2)^2\! - 10^2 \equiv (x-8)(x+12).\:$ Therefore $\rm\: 7^2\mid (x-8)(x+12)\:\Rightarrow\: 7^2\mid x-8\ $ or $\rm\ 7^2\mid x+12\ $ or $\rm\ 7\mid x-8,\, x+12.\:$ The last case is impossible since it implies that $\rm\:mod\ 7\!:\,\ 8\equiv x\equiv -12.$

Quite likely, the first problem is a misprint for $\rm\:f = 2x^\color{#C00}2+7x-4.\:$ Using the AC method, with $\rm\: X = 2x,\:$ we have $\rm\:2f = X^2+7X-8 = (X+8)(X-1) = (2x+8)(2x-1),\:$ therefore $\rm\: f = (x+4)(2x-1).\:$ Now, as above, apply a case analysis to $\rm\ 5^2\!\mid (x+4)(2x-1).$