Collection of exercises on sign analysis for polynomial and rational functions. Step-by-step solutions with sign charts and an analysis of zeros, the domain, and any excluded points.
Sign analysis means determining for which values of the variable a function is positive, negative, or equal to zero. To do this, we study the zeros, the domain, and, when necessary, the sign of each factor.
Exercise 1 — level ★★☆☆☆
\[ f(x) = (x-3)(x+1) \]
Result
\[ f(x) > 0 \text{ for } x < -1 \text{ or } x > 3\]
\[ f(x) = 0 \text{ for } x=-1,\,3\]
\[ f(x) < 0 \text{ for } -1 < x < 3 \]
Solution
Zeros
The zeros are the values that make at least one factor equal to zero:
\[ x+1=0 \Rightarrow x=-1, \qquad x-3=0 \Rightarrow x=3. \]
Sign chart
Conclusion
\(f(x)>0\) for \(x<-1\) or \(x>3\).
\(f(x)=0\) for \(x=-1\) and \(x=3\).
\(f(x)<0\) for \(-1<x<3\).
Exercise 2 — level ★★☆☆☆
\[ f(x) = -(x+2)(x-4) \]
Result
\[ f(x) > 0 \text{ for } -2 < x < 4\]
\[ f(x) = 0 \text{ for } x=-2,\,4\]
\[ f(x) < 0 \text{ for } x < -2 \text{ or } x > 4 \]
Solution
Observation
The factor \(-1\) reverses the sign of the product \((x+2)(x-4)\).
Zeros
\[ x+2=0 \Rightarrow x=-2, \qquad x-4=0 \Rightarrow x=4. \]
Sign chart
Conclusion
\(f(x)>0\) for \(-2<x<4\).
\(f(x)=0\) for \(x=-2\) and \(x=4\).
\(f(x)<0\) for \(x<-2\) or \(x>4\).
Exercise 3 — level ★★☆☆☆
\[ f(x) = x^2 - 5x + 6 \]
Result
\[ f(x) > 0 \text{ for } x < 2 \text{ or } x > 3\]
\[ f(x) = 0 \text{ for } x=2,\,3\]
\[ f(x) < 0 \text{ for } 2 < x < 3 \]
Solution
Factorization
\[ x^2 - 5x + 6 = (x-2)(x-3) \]
Zeros
\(x=2\) and \(x=3\).
Sign chart
Conclusion
\(f(x)>0\) for \(x<2\) or \(x>3\).
\(f(x)<0\) for \(2<x<3\).
Exercise 4 — level ★★☆☆☆
\[ f(x) = -x^2 + x + 6 \]
Result
\[ f(x) > 0 \text{ for } -2 < x < 3\]
\[ f(x) = 0 \text{ for } x=-2,\,3\]
\[ f(x) < 0 \text{ for } x < -2 \text{ or } x > 3 \]
Solution
Factorization
\[ -x^2+x+6 = -(x^2-x-6)=-(x-3)(x+2) \]
Zeros
\(x=-2\) and \(x=3\).
Sign chart
Conclusion
\(f(x)>0\) for \(-2<x<3\).
\(f(x)<0\) for \(x<-2\) or \(x>3\).
Exercise 5 — level ★★★☆☆
\[ f(x) = (x+1)(x-2)(x-5) \]
Result
\[ f(x) > 0 \text{ for } -1 < x < 2 \text{ or } x > 5\]
\[ f(x) = 0 \text{ for } x=-1,\,2,\,5\]
\[ f(x) < 0 \text{ for } x < -1 \text{ or } 2 < x < 5 \]
Solution
Zeros
\(x=-1\), \(x=2\), \(x=5\).
Observation
The zeros are simple, so the sign of the product changes when crossing each of them.
Sign chart
Conclusion
\(f(x)>0\) for \(-1<x<2\) or \(x>5\).
\(f(x)<0\) for \(x<-1\) or \(2<x<5\).
Exercise 6 — level ★★★☆☆
\[ f(x) = \frac{x-1}{x+3} \]
Result
\[ f(x) > 0 \text{ for } x < -3 \text{ or } x > 1\]
\[ f(x)=0 \text{ for } x=1\]
\[ f(x) < 0 \text{ for } -3 < x < 1 \]
Solution
Domain
The denominator is zero at \(x=-3\). Therefore \(x\neq -3\).
Zero
The numerator is zero at \(x=1\).
Sign chart
Conclusion
\(x=-3\) is excluded from the domain; \(x=1\) is a zero of the function.
Exercise 7 — level ★★★☆☆
\[ f(x) = x^3 - 4x \]
Result
\[ f(x) > 0 \text{ for } -2 < x < 0 \text{ or } x > 2\]
\[ f(x)=0 \text{ for } x=-2,\,0,\,2\]
\[ f(x) < 0 \text{ for } x < -2 \text{ or } 0 < x < 2 \]
Solution
Factorization
\[ x^3-4x=x(x^2-4)=x(x-2)(x+2) \]
Zeros
\(x=-2\), \(x=0\), \(x=2\).
Sign chart
Exercise 8 — level ★★★☆☆
\[ f(x) = \frac{x^2-4}{x-1} \]
Result
\[ f(x) > 0 \text{ for } -2 < x < 1 \text{ or } x > 2\]
\[ f(x)=0 \text{ for } x=-2,\,2\]
\[ f(x) < 0 \text{ for } x < -2 \text{ or } 1 < x < 2 \]
Solution
Factorization
\[ f(x)=\frac{(x-2)(x+2)}{x-1} \]
Domain
\(x\neq 1\).
Sign chart
Exercise 9 — level ★★★☆☆
\[ f(x) = (2x+1)(x-3)^2 \]
Result
\[ f(x) > 0 \text{ for } x > -\frac{1}{2},\ x\neq 3\]
\[ f(x)=0 \text{ for } x=-\frac{1}{2},\,3\]
\[ f(x) < 0 \text{ for } x < -\frac{1}{2} \]
Solution
Zeros
\(x=-\frac{1}{2}\) is a simple zero; \(x=3\) is a double zero.
Observation
The factor \((x-3)^2\) is always non-negative: the sign does not change when crossing \(x=3\).
Sign chart
Exercise 10 — level ★★★☆☆
\[ f(x) = \frac{(x+2)(x-1)}{x(x-4)} \]
Result
\[ f(x) > 0 \text{ for } x < -2 \text{ or } 0 < x < 1 \text{ or } x > 4\]
\[ f(x) < 0 \text{ for } -2 < x < 0 \text{ or } 1 < x < 4 \]
Solution
Domain
\(x\neq 0\), \(x\neq 4\).
Zeros
\(x=-2\) and \(x=1\).
Sign chart
Exercise 11 — level ★★★★☆
\[ f(x) = x^4 - 5x^2 + 4 \]
Result
\[ f(x) > 0 \text{ for } x < -2,\ -1 < x < 1,\ x > 2\]
\[ f(x) < 0 \text{ for } -2 < x < -1 \text{ or } 1 < x < 2 \]
Solution
Factorization
\[ x^4-5x^2+4=(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2) \]
Zeros
\(x=-2\), \(x=-1\), \(x=1\), \(x=2\).
Sign chart
Exercise 12 — level ★★★★☆
\[ f(x) = \frac{x^2-3x+2}{x^2+x-6} \]
Result
\[ f(x) > 0 \text{ for } x < -3 \text{ or } 1 < x < 2 \text{ or } x > 2\]
\[ f(x) < 0 \text{ for } -3 < x < 1 \]
Solution
Factorization
\[ f(x)=\frac{(x-1)(x-2)}{(x+3)(x-2)} \]
Domain
\(x\neq -3\), \(x\neq 2\).
Simplification
For \(x\neq 2\), the function has the same sign as \(\frac{x-1}{x+3}\), but the point \(x=2\) remains excluded from the domain.
Sign chart
Exercise 13 — level ★★★★☆
\[ f(x) = (x^2-1)(x^2-9) \]
Result
\[ f(x) > 0 \text{ for } x < -3,\ -1 < x < 1,\ x > 3\]
\[ f(x) < 0 \text{ for } -3 < x < -1 \text{ or } 1 < x < 3 \]
Solution
Factorization
\[ (x^2-1)(x^2-9)=(x-1)(x+1)(x-3)(x+3) \]
Zeros
\(x=-3\), \(x=-1\), \(x=1\), \(x=3\).
Sign chart
Exercise 14 — level ★★★★☆
\[ f(x) = \frac{x^3-x}{x^2-4} \]
Result
\[ f(x) > 0 \text{ for } -2 < x < -1 \text{ or } 0 < x < 1 \text{ or } x > 2\]
\[ f(x) < 0 \text{ for } x < -2 \text{ or } -1 < x < 0 \text{ or } 1 < x < 2 \]
Solution
Factorization
\[ f(x)=\frac{x(x-1)(x+1)}{(x-2)(x+2)} \]
Domain
\(x\neq -2\), \(x\neq 2\).
Zeros
\(x=-1\), \(x=0\), \(x=1\).
Sign chart
Exercise 15 — level ★★★★☆
\[ f(x) = x^2(x-2)(x+3) \]
Result
\[ f(x) > 0 \text{ for } x < -3 \text{ or } x > 2\]
\[ f(x)=0 \text{ for } x=-3,\,0,\,2\]
\[ f(x) < 0 \text{ for } -3 < x < 0 \text{ or } 0 < x < 2 \]
Solution
Zeros
\(x=-3\), \(x=0\), \(x=2\).
Observation
The zero \(x=0\) is double, so the sign does not change when crossing it.
Sign chart
Exercise 16 — level ★★★★★
\[ f(x) = \frac{x^2(x+1)}{(x-1)^2(x+2)} \]
Result
\[ f(x) > 0 \text{ for } x < -2 \text{ or } -1 < x < 0 \text{ or } 0 < x < 1 \text{ or } x > 1\]
\[ f(x) < 0 \text{ for } -2 < x < -1 \]
Solution
Domain
\(x\neq -2\), \(x\neq 1\).
Zeros
\(x=-1\) and \(x=0\).
Observation
The factors \(x^2\) and \((x-1)^2\) do not change sign. The zero \(x=0\) is double and the pole \(x=1\) is double.
Sign chart
Exercise 17 — level ★★★★★
\[ f(x) = \frac{x^4-1}{x^3-x} \]
Result
\[ f(x) > 0 \text{ for } 0 < x < 1 \text{ or } x > 1\]
\[ f(x) < 0 \text{ for } x < -1 \text{ or } -1 < x < 0 \]
Solution
Factorization
\[ x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1) \]
\[ x^3-x=x(x^2-1)=x(x-1)(x+1) \]
Domain
\[ x\neq -1,\quad x\neq 0,\quad x\neq 1 \]
Simplification
\[ f(x)=\frac{(x-1)(x+1)(x^2+1)}{x(x-1)(x+1)}=\frac{x^2+1}{x} \]
Since \(x^2+1>0\) for every \(x\in\mathbb{R}\), the sign depends only on \(x\), but \(x=-1\), \(x=0\), and \(x=1\) remain excluded from the domain.
Sign chart
Exercise 18 — level ★★★★★
\[ f(x) = (x^2+2x-3)^2 \]
Result
\[ f(x)\geq 0 \text{ for every } x\in\mathbb{R}\]
\[ f(x)=0 \text{ for } x=-3 \text{ and } x=1 \]
Solution
Factorization
\[ x^2+2x-3=(x+3)(x-1) \]
\[ f(x)=\big[(x+3)(x-1)\big]^2 \]
Zeros
\(x=-3\) and \(x=1\).
Observation
The function is the square of a real expression, so it is always non-negative.
Sign chart
Conclusion
\(f(x)>0\) for all real numbers except \(-3\) and \(1\).
\(f(x)=0\) for \(x=-3\) and \(x=1\).
Exercise 19 — level ★★★★★
\[ f(x) = \frac{(x+1)^2(x-2)}{x^2(x+3)} \]
Result
\[ f(x) > 0 \text{ for } x < -3 \text{ or } x > 2\]
\[ f(x) < 0 \text{ for } -3 < x < -1 \text{ or } -1 < x < 0 \text{ or } 0 < x < 2 \]
Solution
Domain
\(x\neq -3\), \(x\neq 0\).
Zeros
\(x=-1\) is a double zero; \(x=2\) is a simple zero.
Observation
The factors \((x+1)^2\) and \(x^2\) do not change sign. The sign depends on the factors \(x-2\) and \(x+3\), taking the excluded points into account.
Sign chart
Exercise 20 — level ★★★★★
\[ f(x) = \frac{x^3+x^2-4x-4}{x^2-x-2} \]
Result
\[ f(x) > 0 \text{ for } -2 < x < -1 \text{ or } -1 < x < 2 \text{ or } x > 2\]
\[ f(x)=0 \text{ for } x=-2; \quad f(x) < 0 \text{ for } x < -2 \]
Solution
Factorization of the numerator
\[ x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x+1)(x^2-4)=(x+1)(x-2)(x+2) \]
Factorization of the denominator
\[ x^2-x-2=(x-2)(x+1) \]
Domain
\(x\neq -1\), \(x\neq 2\).
Simplification
For \(x\neq -1\) and \(x\neq 2\):
\[ f(x)=\frac{(x+1)(x-2)(x+2)}{(x-2)(x+1)}=x+2 \]
The function therefore has the same sign as \(x+2\), but the points \(x=-1\) and \(x=2\) remain excluded from the domain.
Sign chart
Conclusion
\(f(x)>0\) for \(-2<x<-1\), \(-1<x<2\) or \(x>2\).
\(f(x)<0\) for \(x<-2\).
\(f(x)=0\) for \(x=-2\).
The points \(x=-1\) and \(x=2\) are excluded from the domain.