From a Polynomial Inequality to a Set of Intervals
A polynomial equation asks where a polynomial equals zero. A polynomial inequality asks a broader question: for which real inputs is its value positive, negative, or possibly zero? Factoring helps answer this because the real zeros divide the number line into intervals on which the sign can be determined.
The goal is not just to list the zeros. The zeros are boundary points between intervals, and the inequality may be true on some intervals but not others. Whether a boundary point itself belongs in the answer depends on whether equality is allowed.
The Basic Method
First move all terms to one side, so the inequality compares one polynomial with zero. Subtracting the same expression from both sides preserves the inequality. Then factor the polynomial if possible, find its real zeros, and order the distinct zeros from least to greatest. These zeros divide the real line into intervals.
On each interval, determine whether the polynomial is positive or negative. A test value from the interval is often the quickest way. The test value must lie strictly inside the interval, not at a zero. Once the sign on each interval is known, keep the intervals that satisfy the inequality. Finally, decide whether to include the zeros themselves.
Write it as \(P(x)>0\), \(P(x)\geq 0\), \(P(x)<0\), or \(P(x)\leq 0\).
Factor \(P(x)\) and list each distinct real zero once, from least to greatest.
Choose a test value in each interval and evaluate the sign of \(P(x)\), or use the signs of its factors.
Keep intervals with the required sign, then include a zero exactly when the inequality allows equality.
This approach uses the sign-chart ideas from the earlier material on Polynomial Sign Charts and Interval Signs. It is useful to keep the tasks separate: zeros identify the interval boundaries, interval signs identify where the inequality holds, and the inequality symbol determines whether boundaries are included.
Why the Sign Pattern Works
Suppose a polynomial is written as a nonzero constant times real linear factors, with repeated factors shown according to their multiplicities. Write its distinct real zeros as \(r_1<r_2<\cdots<r_k\), and let the multiplicity of \(r_j\) be \(m_j\). Then the polynomial has the form
Between two consecutive zeros, each factor \(x-r_j\) has a fixed sign: the input \(x\) stays on the same side of \(r_j\) throughout that interval. Therefore, the product also has a fixed sign there. When moving across a zero \(r_j\), the factor \((x-r_j)^{m_j}\) changes sign if \(m_j\) is odd, and keeps its sign if \(m_j\) is even. All the other factors keep their signs in a sufficiently small move across that zero.
This explains why a sign chart needs only one sign determination to start: the signs at the remaining intervals can be tracked by considering the multiplicity of each zero. Direct test values are also valid and can be used to check the pattern. If the polynomial is not fully factored into real linear factors, use a factorization whose remaining factors have known signs, or test values in each interval.
Worked Examples
Worked Example: A Quadratic Inequality with Included Endpoints
Solve \((x-3)(x+1)\leq 0\).
The zeros are found by setting each factor equal to zero:
Ordered from least to greatest, the zeros are \(-1\) and \(3\). They divide the real line into \((-\infty,-1)\), \((-1,3)\), and \((3,\infty)\).
Test \(x=-2\) in the first interval: \((-2-3)(-2+1)=(-5)(-1)=5>0\). Test \(x=0\) in the middle interval: \((0-3)(0+1)=(-3)(1)=-3<0\). Test \(x=4\) in the last interval: \((4-3)(4+1)=(1)(5)=5>0\). Thus the expression is nonpositive on the middle interval and equals zero at both boundary values.
Because the inequality includes equality, include \(x=-1\) and \(x=3\). The answer is \([-1,3]\). Checking the boundaries gives \((-1-3)(-1+1)=(-4)(0)=0\) and \((3-3)(3+1)=(0)(4)=0\), as required.
Worked Example: A Repeated Zero and a Strict Inequality
Solve \(x^2(x-2)(x+1)>0\).
The zeros are \(x=-1\), \(x=0\), and \(x=2\). The zero at \(0\) has multiplicity \(2\), because its factor is \(x^2\); the other zeros have multiplicity \(1\). The ordered zeros divide the real line into four intervals.
Test \(x=-2\): \((-2)^2(-2-2)(-2+1)=4(-4)(-1)=16>0\). Test \(x=-\frac12\): \(\left(-\frac12\right)^2\left(-\frac12-2\right)\left(-\frac12+1\right)=\frac14\left(-\frac52\right)\left(\frac12\right)=-\frac{5}{16}<0\). Test \(x=1\): \(1^2(1-2)(1+1)=1(-1)(2)=-2<0\). Test \(x=3\): \(3^2(3-2)(3+1)=9(1)(4)=36>0\).
The polynomial is positive on \((-\infty,-1)\) and \((2,\infty)\). Since the inequality is strict, none of the zeros is included, including the double zero at \(0\). The solution is \((-\infty,-1)\cup(2,\infty)\).
At each boundary, a factor is zero: the product is \(0\) at \(-1\), at \(0\), and at \(2\). This confirms that these points do not satisfy the strict inequality.
Worked Example: Factoring an Expanded Polynomial First
Solve \(x^3-4x^2-x+4\leq 0\).
Factor by grouping:
Multiplying back verifies the factorization: \((x-1)(x+1)=x^2-1\), and \((x^2-1)(x-4)=x^3-4x^2-x+4\). The zeros are \(-1\), \(1\), and \(4\), giving the intervals \((-\infty,-1)\), \((-1,1)\), \((1,4)\), and \((4,\infty)\).
Use \(x=-2\): \((-3)(-1)(-6)=-18<0\). Use \(x=0\): \((-1)(1)(-4)=4>0\). Use \(x=2\): \((1)(3)(-2)=-6<0\). Use \(x=5\): \((4)(6)(1)=24>0\). The polynomial is nonpositive on the first and third intervals, and it equals zero at all three roots.
Since equality is allowed, include the zeros \(-1\), \(1\), and \(4\). Combining the selected intervals with their included endpoints gives \((-\infty,-1]\cup[1,4]\). Substitution checks the boundary values: the factored form has a zero factor at each of \(-1\), \(1\), and \(4\).
Boundary Values and Inequality Symbols
A zero is always a point where the polynomial’s value is exactly zero. Whether that point belongs in the answer depends on the comparison being solved:
| Inequality | Include a zero of \(P(x)\)? |
|---|---|
| \(P(x)>0\) | No |
| \(P(x)<0\) | No |
| \(P(x)\geq 0\) | Yes |
| \(P(x)\leq 0\) | Yes |
In interval notation, parentheses exclude an endpoint and square brackets include it. Strict inequalities use parentheses at zeros. Non-strict inequalities include every zero, although a zero may be isolated in the solution set if the polynomial has the required sign on neither adjacent interval. For example, an even-multiplicity zero may be included for a non-strict inequality even when the sign is negative on both sides.
Common Mistake: Treating Every Zero as a Sign Change
Consider \((x-2)^2(x+1)>0\). A common error is to assume that the sign changes at every zero, and then mark the polynomial as negative on \((-1,2)\). But \(x=2\) is a double zero, so crossing it does not reverse the sign. A direct test exposes the error: at \(x=0\), which lies in \((-1,2)\), the value is \((0-2)^2(0+1)=4>0\), not negative.
The factor \(x+1\) is negative for \(x<-1\) and positive for \(x>-1\). The squared factor is positive except at \(x=2\), where it is zero. Thus the expression is positive on \((-1,2)\) and \((2,\infty)\), and zero at \(-1\) and \(2\). The strict inequality excludes both zeros, so the solution is \((-1,2)\cup(2,\infty)\). The double zero does not change the sign, but it still fails the strict inequality because the polynomial equals zero there.
Why This Method Is Useful
Sign analysis turns an inequality into a precise description of all the real inputs that satisfy it. It also guards against a tempting but unreliable shortcut: looking only at the zeros. Knowing the zeros does not tell you which intervals have positive values and which have negative values. Factoring, checking interval signs, and applying the endpoint rule together provide that information.
Before finalizing a solution, check that the real zeros are listed in order, that each test value belongs to its stated interval, and that the sign of every product is correct. Then check the boundary rule separately. This last check is especially important: strict and non-strict inequalities can have the same solution intervals between zeros but differ at their endpoints.
Check Your Understanding
For each problem, factor if needed, determine the signs on the intervals, and state the solution set with correct endpoint notation.
- Solve \((x+2)(x-1)>0\). Which intervals satisfy the inequality, and are either zero included?
- Solve \((x-3)^2(x+1)\leq 0\). What happens to the sign as you cross the double zero?
- For \(x^2-5x+6\geq 0\), find the zeros, order them, and write the solution set.
- Explain why a zero of a polynomial must be excluded from the solution of a strict inequality involving \(P(x)>0\) or \(P(x)<0\).
