Exercise 1829Algebra › PolynomialsL1Added 3 Nov 2026

Roots of a cubic and discriminant

#cubic#discriminant#roots

Statement

Find the number of real roots of $P(X)=X^3-3X+1$ as a function of nothing (it is fixed), using the discriminant.

Hint

The discriminant of $X^3+pX+q$ is $\Delta=-4p^3-27q^2$ .

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Solution

Discriminant. With $p=-3$ , $q=1$ : $\Delta=-4(-27)-27(1)=108-27=81>0$ .

Conclusion. Since $\Delta>0$ , the cubic has three distinct real roots.

Verification. $P(-2)=-8+6+1=-1<0$ , $P(-1)=-1+3+1=3>0$ , $P(0)=1>0$ , $P(1)=1-3+1=-1<0$ , $P(2)=8-6+1=3>0$ . So roots in $]-2,-1[$ , $]0,1[$ , and $]1,2[$ .

Roots of a cubic and discriminant

Statement

Solution

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