Exercise 27Calculus › Single-variable calculus › Continuity › Intermediate Value TheoremHigh schoolAdded 2 Nov 2026

Locating the roots of a cubic

#intermediate value theorem#polynomial roots#bracketing

Statement

Show that the equation $x^3+x^2-4x+1=0$ has exactly three real solutions, and find three disjoint intervals of length $1$ containing them.

Hint

Evaluate $P(x)=x^3+x^2-4x+1$ at the integers $-3,-2,0,1,2$ and look at the signs.

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Solution

Values. $P(-3)=-5$ , $P(-2)=5$ , $P(0)=1$ , $P(1)=-1$ , $P(2)=5$ .

Applying the intermediate value theorem. $P$ is continuous, and its sign changes:
- between $-3$ and $-2$ ( $-5<0<5$ ): a root in $(-3,-2)$ ,
- between $0$ and $1$ ( $1>0>-1$ ): a root in $(0,1)$ ,
- between $1$ and $2$ ( $-1<0<5$ ): a root in $(1,2)$ .

These three intervals are disjoint, so we have found three distinct roots.

No more roots. A polynomial of degree $3$ has at most $3$ roots. Hence there are exactly three, and each of the intervals contains exactly one.

Numerical values. The roots are approximately $-2.651$ , $0.274$ and $1.377$ .

Locating the roots of a cubic

Statement

Solution

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