Exercise 947Algebra › Equations and radicalsL1Added 3 Nov 2026

A quartic by factoring

#quartic#factoring#roots

Statement

Solve in $\mathbb{R}$ the equation
$x^4-5x^3+5x^2+5x-6=0.$

Hint

Test $x=1$ and $x=-1$ .

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Test roots. $x=1$ : $1-5+5+5-6=0$ . Root. $x=-1$ : $1+5+5-5-6=0$ . Root.

Factor. Divide by $(x-1)(x+1)=x^2-1$ :
$x^4-5x^3+5x^2+5x-6=(x^2-1)(x^2-5x+6)=(x-1)(x+1)(x-2)(x-3).$

Roots. $x\in\{-1,1,2,3\}$ .

Conclusion. $x\in\{-1,1,2,3\}$ .