Sign charts: where an expression is positive or negative

A fictional workshop models daily profit by $P(q)=-10(q-2)(q-8)$ , where $q$ is an integer number of items from 0 to 10 and $P$ is in currency units. When is profit positive? Finding only the break-even values 2 and 8 is insufficient: we need the intervals between them.

Use factoring, quadratic equations and inequalities. This invented model applies only on its stated domain.

Why one test value works within an interval

The factor $q-2$ changes sign only at 2; $q-8$ only at 8. Left of 2 both are negative, between 2 and 8 their signs differ, and right of 8 both are positive. Their product has signs positive, negative, positive. The factor −10 reverses them: profit is positive for $2<q<8$ , hence integers 3,4,5,6,7. At $q=5$ , profit is 90; at 2 and 8 it is zero.

Zeros of factors and forbidden denominator values divide the line into intervals. Each factor keeps its sign inside each interval, so the product or quotient does too. A test value identifies the sign; it does not by itself prove that the sign cannot change. That follows from the expression’s structure. Rational inequalities, OpenStax §7.6.

A zero and a pole play different roles

For real $x$ , solve:

\frac{(x-1)^2(x+2)}{x-3}\ge0.

First $x\ne3$ . Numerator zeros are −2 and 1. At 3 the denominator is zero and the numerator is not: this is a pole, excluded even for a non-strict inequality. At 1 the squared factor has even multiplicity and is positive on both sides. The sign therefore does not change there.

Quotient signs

R(x) = (x − 1)²(x + 2)/(x − 3)

Interval	Sign	R(x) ≥ 0
(−∞, −2)	+	✓
(−2, 1)	−	—
(1, 3)	−	—
(3, ∞)	+	✓

Points: R(−2) = 0; R(1) = 0; x ≠ 3.

−2: zero; 1: double zero; 3: pole, excluded. Even multiplicity at 1 preserves the sign. For ≥0 include zeros, never the pole.

The interval signs are $+,-,-,+$ . For $\ge0$ , select positive intervals and admissible zeros:

(-\infty,-2]\cup\{1\}\cup(3,\infty).

One is an isolated solution: neighboring values are negative, but the value at 1 is zero. For strict $>0$ , neither −2 nor 1 would be included. Blindly alternating signs misses this feature. An odd power changes a factor’s sign across its zero; an even power does not. Root multiplicity, OpenStax §5.3.

Original exclusions survive cancellation. Not every forbidden point is a pole: $\frac{x-1}{x-1}$ has a removable hole at 1, but 1 still cannot be included.

Branch · A radical inequalityDomain and signs first, then intervals.

When may we square both sides?

In $\sqrt{x+2}>x$ , require $x\ge-2$ . On $[-2,0)$ , the left side is nonnegative and the right negative, so the inequality holds automatically. For $x\ge0$ , both sides are nonnegative and squaring is equivalent: $x+2>x^2$ , or $(x-2)(x+1)<0$ . Thus $-1<x<2$ ; intersecting with this case gives $0\le x<2$ . The union is $[-2,2)$ . The endpoint −2 works, $0>-2$ ; 2 fails, since $2>2$ is false.

Keep three separate pieces of information: signs on open intervals, values at zeros, and forbidden denominator values. Together they produce the complete answer.