Feb 24, 2014 Thread starter #1 S suvadip Member Feb 21, 2013 69 Show that 2x^3+x^2+3x-1 = 0 (mod 5) has exactly three solutions How to proceed with it? Last edited: Feb 24, 2014
Feb 25, 2014 #3 S soroban Well-known member Feb 2, 2012 409 Hello, suvadip! I suppose someone wants to see all the steps. [tex]\text{Show that }\,2x^3+x^2+3x-1\:\equiv\:0\text{ (mod 5)}[/tex] [tex]\text{has exactly three solutions.}[/tex] Click to expand... We have: .[tex]2x^3+x^2 + 3x-1\:\equiv\:0\text{ (mod 5)}[/tex] Then: .[tex]2x^3 + x^2 - 2x - 1 \:\equiv\:0\text{ (mod 5)}[/tex] Factor: .[tex]x^2(2x+1) - (2x+1) \:\equiv\;0\text{ (mod 5)}[/tex] Factor: .[tex](2x+1)(x^2-1) \:\equiv\:0\text{ (mod 5)}[/tex] Factor: .[tex](2x+1)(x-1)(x+1) \:\equiv\:0\text{ (mod 5)}[/tex] [tex]2x+1\:\equiv\:0 \text{ (mod 5)}\quad\Rightarrow\quad 2x \:\equiv\:-1 \text{ (mod 5)}[/tex] . . [tex]2x \:\equiv\:4 \text{ (mod 5)} \quad\Rightarrow\quad \boxed{x \:\equiv\:2\text{ (mod 5)}} [/tex] [tex]x-1\:\equiv\:0\text{ (mod 5)} \quad\Rightarrow\quad \boxed{x \:\equiv\:1\text{ (mod 5)}}[/tex] [tex]x+1\:\equiv\:0\text{ (mod 5)} \quad\Rightarrow\quad x \:\equiv\:-1\text{ (mod 5)}[/tex] . . [tex]\boxed{x\:\equiv\:4\text{ (mod 5)}}[/tex]
Hello, suvadip! I suppose someone wants to see all the steps. [tex]\text{Show that }\,2x^3+x^2+3x-1\:\equiv\:0\text{ (mod 5)}[/tex] [tex]\text{has exactly three solutions.}[/tex] Click to expand... We have: .[tex]2x^3+x^2 + 3x-1\:\equiv\:0\text{ (mod 5)}[/tex] Then: .[tex]2x^3 + x^2 - 2x - 1 \:\equiv\:0\text{ (mod 5)}[/tex] Factor: .[tex]x^2(2x+1) - (2x+1) \:\equiv\;0\text{ (mod 5)}[/tex] Factor: .[tex](2x+1)(x^2-1) \:\equiv\:0\text{ (mod 5)}[/tex] Factor: .[tex](2x+1)(x-1)(x+1) \:\equiv\:0\text{ (mod 5)}[/tex] [tex]2x+1\:\equiv\:0 \text{ (mod 5)}\quad\Rightarrow\quad 2x \:\equiv\:-1 \text{ (mod 5)}[/tex] . . [tex]2x \:\equiv\:4 \text{ (mod 5)} \quad\Rightarrow\quad \boxed{x \:\equiv\:2\text{ (mod 5)}} [/tex] [tex]x-1\:\equiv\:0\text{ (mod 5)} \quad\Rightarrow\quad \boxed{x \:\equiv\:1\text{ (mod 5)}}[/tex] [tex]x+1\:\equiv\:0\text{ (mod 5)} \quad\Rightarrow\quad x \:\equiv\:-1\text{ (mod 5)}[/tex] . . [tex]\boxed{x\:\equiv\:4\text{ (mod 5)}}[/tex]