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Exercise 633Algebra › PolynomialsM1Added 5 Nov 2026

How many real roots does x⁵ − 5x + a have?

#polynomials#derivative#number of roots#parameter

Statement

For a real number aa, determine the number of distinct real roots of Pa(x)=x5−5x+aP_a(x)=x^5-5x+a.

Hint

Pa′(x)=5(x4−1)P_a'(x)=5(x^4-1). A polynomial is monotone between consecutive critical points, so it has at most one root in each such interval.

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Solution

Critical points. Pa′(x)=5(x2−1)(x2+1)P_a'(x)=5(x^2-1)(x^2+1) vanishes only at x=±1x=\pm1. Hence PaP_a is increasing on (−∞,−1](-\infty,-1], decreasing on [−1,1][-1,1] and increasing on [1,+∞)[1,+\infty).

Values. Pa(−1)=4+aP_a(-1)=4+a (local maximum) and Pa(1)=a−4P_a(1)=a-4 (local minimum). Moreover Pa(x)→−∞P_a(x)\to-\infty as x→−∞x\to-\infty and Pa(x)→+∞P_a(x)\to+\infty as x→+∞x\to+\infty.

Counting. On each of the three monotone pieces there is at most one root, and by the intermediate value theorem there is one exactly when the endpoint values (or limits) have opposite signs or one is zero.
- ∣a∣<4|a|<4: Pa(−1)>0>Pa(1)P_a(-1)>0>P_a(1). One root in each piece: three roots.
- a=4a=4: Pa(−1)=8>0P_a(-1)=8>0 and Pa(1)=0P_a(1)=0. The root x=1x=1 is a double root and there is one more root on (−∞,−1)(-\infty,-1): two distinct roots.
- a=−4a=-4: symmetric: Pa(−1)=0P_a(-1)=0 is a double root and there is one root on (1,∞)(1,\infty): two distinct roots.
- a>4a>4: Pa(1)>0P_a(1)>0, so Pa>0P_a>0 on [−1,∞)[-1,\infty). One root, on (−∞,−1)(-\infty,-1): one root.
- a<−4a<-4: symmetric, one root on (1,∞)(1,\infty).

Lesson. To count the real roots of a polynomial depending on a parameter, study the critical values: roots appear or disappear exactly when a critical value crosses 00.