Porisma

Worked solution

Check a repeated root and include isolated equality solutions

Solve (i) (x−1)(x+1)(x+3)2>0(x-1)(x+1)(x+3)^2>0 and (ii) (x−1)(x+1)(x+3)2≤0(x-1)(x+1)(x+3)^2\le0.

CJC · Lecture · Example 2

Graphical method: read crossings and a touch

The polynomial has roots −3,−1,1. At −3 the squared factor makes the graph touch the axis without changing sign; the other two roots cross. The sketch is positive on the two intervals left of −1 except at −3, negative between −1 and 1, and positive beyond 1.

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Algebraic method: retain both endpoint conventions

Test signs on (−∞,−3),(−3,−1),(−1,1),(1,∞)(-\infty,-3),(-3,-1),(-1,1),(1,\infty): +,+,−,+. For (i), exclude all zeros: x<−3 or −3<x<−1 or x>1. For (ii), include the zeros as well as the negative interval: x=−3 or −1≤x≤1.

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Apply the source remarks

For ≥ or ≤, check every critical value carefully. Do not assume signs always alternate: a repeated root with even multiplicity preserves the sign, as at −3. The source contrasts this with Example 1’s simple roots; more generally odd multiplicity changes sign and even multiplicity does not.

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Compare your sketch

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The source quartic touches at −3 and crosses at −1 and 1. Two sign lines retain open endpoints for the positive case and filled roots for the nonpositive case;−3 is an isolated nonpositive solution.
The source quartic touches at −3 and crosses at −1 and 1. Two sign lines retain open endpoints for the positive case and filled roots for the nonpositive case;−3 is an isolated nonpositive solution.
The source quartic touches at −3 and crosses at −1 and 1. Two sign lines retain open endpoints for the positive case and filled roots for the nonpositive case;−3 is an isolated nonpositive solution.

Common mistakes

For the nonpositive inequality, −3 is a valid isolated point even though both neighbouring intervals are positive.

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