Therefore, the solutions are x = -1, 1, and 2 (mod 5).

[Suvadip]

Show that 2x^3+x^2+3x-1 = 0 (mod 5)
has exactly three solutionsHow to proceed with it?

 

[ThePerfectHacker]

Substitute $x=0,1,2,3,4$.

 

[soroban]

Hello, suvadip!

I suppose someone wants to see all the steps.

$$\text{Show that }\,2x^3+x^2+3x-1\:\equiv\:0\text{ (mod 5)}$$
$$\text{has exactly three solutions.}$$

We have: .$$2x^3+x^2 + 3x-1\:\equiv\:0\text{ (mod 5)}$$

Then: .$$2x^3 + x^2 - 2x - 1 \:\equiv\:0\text{ (mod 5)}$$

Factor: .$$x^2(2x+1) - (2x+1) \:\equiv\;0\text{ (mod 5)}$$

Factor: .$$(2x+1)(x^2-1) \:\equiv\:0\text{ (mod 5)}$$

Factor: .$$(2x+1)(x-1)(x+1) \:\equiv\:0\text{ (mod 5)}$$$$2x+1\:\equiv\:0 \text{ (mod 5)}\quad\Rightarrow\quad 2x \:\equiv\:-1 \text{ (mod 5)}$$

. . $$2x \:\equiv\:4 \text{ (mod 5)} \quad\Rightarrow\quad \boxed{x \:\equiv\:2\text{ (mod 5)}}$$$$x-1\:\equiv\:0\text{ (mod 5)} \quad\Rightarrow\quad \boxed{x \:\equiv\:1\text{ (mod 5)}}$$$$x+1\:\equiv\:0\text{ (mod 5)} \quad\Rightarrow\quad x \:\equiv\:-1\text{ (mod 5)}$$

. . $$\boxed{x\:\equiv\:4\text{ (mod 5)}}$$