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

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The equation 2x^3 + x^2 + 3x - 1 ≡ 0 (mod 5) has been shown to have exactly three solutions: x ≡ -1, 1, and 2 (mod 5). The process involved substituting values for x from 0 to 4 and factoring the polynomial. The factors were simplified to (2x + 1)(x - 1)(x + 1) ≡ 0 (mod 5). This led to the identification of the three distinct solutions. The discussion concludes that these solutions are valid within the modular arithmetic framework.
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Show that 2x^3+x^2+3x-1 = 0 (mod 5)
has exactly three solutionsHow to proceed with it?
 
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Substitute $x=0,1,2,3,4$.
 
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)}}
 
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