Why Organize Polynomial Signs?
A factored polynomial reveals where its value is zero, but its factors also tell us whether the polynomial is positive or negative between those zeros. A sign chart records that information in an organized way: it lists the real zeros in order and gives the polynomial’s sign on each interval they determine.
This is useful because a polynomial’s sign can change as the input moves along the real line, but it cannot change arbitrarily. Its real zeros divide the line into intervals, and the factor multiplicities indicate whether the sign changes when passing through a zero. In this tutorial, the goal is to construct and check the chart itself.
Zeros, Multiplicities, and Intervals
Suppose a nonzero polynomial is written in factored form as \(P(x)=a\prod_{i=1}^{k}(x-r_i)^{m_i}\), where \(a\ne 0\), the \(r_i\) are distinct real numbers, and each \(m_i\) is a positive integer. Then \(r_i\) is a zero of multiplicity \(m_i\). A factor such as \((x-r_i)^{m_i}\) is zero at \(x=r_i\), so the polynomial is zero there.
List the distinct real zeros in increasing order, say \(r_1<r_2<\cdots<r_k\). They divide the real line into \(k+1\) open intervals: \((-\infty,r_1)\), \((r_1,r_2)\), and so on, ending with \((r_k,\infty)\). The polynomial has no zero inside any one of these intervals.
To determine its sign on an interval, choose a convenient test input from that interval and evaluate the factors. The test input must lie strictly inside the interval; choosing an endpoint would give a zero and would not reveal the sign on the interval.
Why the Sign Stays Constant Between Zeros
A polynomial is continuous. If it took a positive value at one point of an interval and a negative value at another, the Intermediate Value Theorem would imply that it took the value zero somewhere between those points. But an interval between consecutive real zeros contains no zero of the polynomial. Therefore, the polynomial’s sign is constant throughout that interval.
This gives the justification for testing just one input: once the sign is determined at that input, it is the sign everywhere in the same interval. The test is not a guess about nearby values; it represents the entire interval because there are no zeros within it.
How Multiplicity Predicts a Sign Change
Multiplicity provides a second way to check the chart. Near a zero \(c\) of multiplicity \(m\), write \(P(x)=(x-c)^mG(x)\), where \(G(c)\ne 0\). Since \(G\) is continuous and nonzero at \(c\), it has the same sign just to the left and right of \(c\), provided those points are sufficiently close to \(c\). The sign difference, if any, comes from \((x-c)^m\).
On the left of \(c\), \(x-c\) is negative; on the right, it is positive. If \(m\) is even, raising either value to the power \(m\) gives a positive value, so the factor’s sign is the same on both sides. If \(m\) is odd, the power is negative on the left and positive on the right, so its sign reverses. Thus an odd-multiplicity zero changes the polynomial’s sign, while an even-multiplicity zero does not.
This rule is a valuable check, not a substitute for understanding the factors and the overall sign. In particular, the sign on the far-right interval depends on the leading coefficient and the signs of all factors there.
A Reliable Construction Process
Read them from the real linear factors, then list distinct zeros from least to greatest. Record each multiplicity.
Use the ordered zeros as boundaries to form the open intervals on the real line.
Choose an easy value inside each interval and determine the sign of every factor and the leading constant.
Across odd multiplicities, neighboring interval signs must differ; across even multiplicities, they must agree.
Worked Example: A Polynomial with Three Real Zeros
Worked Example: A Polynomial with Three Real Zeros
Construct a sign chart for \(q(x)=-2(x+2)^3(x-1)(x-3)^2\).
The real zeros are \(x=-2\), \(x=1\), and \(x=3\), with multiplicities \(3\), \(1\), and \(2\), respectively. In increasing order, they give the intervals \((-\infty,-2)\), \((-2,1)\), \((1,3)\), and \((3,\infty)\).
Choose \(x=-3\) in the first interval. The factors have values \(x+2=-1\), \(x-1=-4\), and \(x-3=-6\), so \((x+2)^3=-1\) and \((x-3)^2=36\). Therefore \(q(-3)=(-2)(-1)(-4)(36)=-288\), which is negative.
Choose \(x=0\) in \((-2,1)\). Then \(x+2=2\), \(x-1=-1\), and \(x-3=-3\), giving \(q(0)=(-2)(2^3)(-1)((-3)^2)=(-2)(8)(-1)(9)=144\), which is positive.
Choose \(x=2\) in \((1,3)\). Then \(x+2=4\), \(x-1=1\), and \(x-3=-1\), giving \(q(2)=(-2)(4^3)(1)((-1)^2)=(-2)(64)(1)(1)=-128\), which is negative.
Choose \(x=4\) in \((3,\infty)\). Then \(x+2=6\), \(x-1=3\), and \(x-3=1\), giving \(q(4)=(-2)(6^3)(3)(1^2)=(-2)(216)(3)=-1296\), which is negative.
The chart is therefore:
| Interval | Test input | Polynomial value | Sign |
|---|---|---|---|
| \((-\infty,-2)\) | \(-3\) | \(q(-3)=-288\) | Negative |
| \((-2,1)\) | \(0\) | \(q(0)=144\) | Positive |
| \((1,3)\) | \(2\) | \(q(2)=-128\) | Negative |
| \((3,\infty)\) | \(4\) | \(q(4)=-1296\) | Negative |
Check the transitions: the sign changes at \(-2\), whose multiplicity is odd, and at \(1\), whose multiplicity is odd. It does not change at \(3\), whose multiplicity is even. These checks agree with the tested signs.
Worked Example: Checking Signs from Multiplicity
Worked Example: Checking Signs from Multiplicity
Construct a sign chart for \(p(x)=3(x+3)^2x(x-4)^3\).
The ordered zeros are \(-3\), \(0\), and \(4\), with multiplicities \(2\), \(1\), and \(3\). The intervals are \((-\infty,-3)\), \((-3,0)\), \((0,4)\), and \((4,\infty)\).
For \(x=-4\), in the first interval, the factors \(x+3\), \(x\), and \(x-4\) are \(-1\), \(-4\), and \(-8\). Thus \(p(-4)=3(-1)^2(-4)(-8)^3=3(1)(-4)(-512)=6144\), positive.
For \(x=-1\), in the second interval, the factors are \(2\), \(-1\), and \(-5\). Thus \(p(-1)=3(2)^2(-1)(-5)^3=3(4)(-1)(-125)=1500\), positive.
For \(x=1\), in the third interval, the factors are \(4\), \(1\), and \(-3\). Thus \(p(1)=3(4)^2(1)(-3)^3=3(16)(1)(-27)=-1296\), negative.
For \(x=5\), in the last interval, the factors are \(8\), \(5\), and \(1\). Thus \(p(5)=3(8)^2(5)(1)^3=3(64)(5)=960\), positive.
The signs, in interval order, are positive, positive, negative, positive. The sign stays the same across \(-3\), of even multiplicity, and changes across both \(0\) and \(4\), of odd multiplicity. The test-point calculations and multiplicity check agree.
Worked Example: A Factor with No Real Zeros
Worked Example: A Factor with No Real Zeros
Construct a sign chart for \(r(x)=-(x-2)^2(x+1)(x^2+4)\).
The factor \(x^2+4\) is positive for every real \(x\), because \(x^2\ge 0\) implies \(x^2+4\ge 4>0\). It therefore contributes no real zero and no interval boundary. The real zeros are \(-1\), of multiplicity \(1\), and \(2\), of multiplicity \(2\). The intervals are \((-\infty,-1)\), \((-1,2)\), and \((2,\infty)\).
At \(x=-2\), the factors \((x-2)^2\), \(x+1\), and \(x^2+4\) are \(16\), \(-1\), and \(8\). Hence \(r(-2)=-(16)(-1)(8)=128\), positive.
At \(x=0\), those factors are \(4\), \(1\), and \(4\). Hence \(r(0)=-(4)(1)(4)=-16\), negative.
At \(x=3\), they are \(1\), \(4\), and \(13\). Hence \(r(3)=-(1)(4)(13)=-52\), negative.
The chart signs are positive, negative, and negative. The sign changes across \(-1\), which has odd multiplicity, and stays the same across \(2\), which has even multiplicity. The factor \(x^2+4\) never vanishes, so it does not split any interval.
A Common Mistake: Misreading a Factor Value
When evaluating a test input, substitute it into each factor before applying the exponent. For the first example, at \(x=-3\), the factor \(x+2\) is \(-3+2=-1\), so \((x+2)^3=(-1)^3=-1\). It is not \((-2)^3\); the exponent applies to the value of the factor after substitution.
A mistaken calculation might write \(q(-3)=(-2)(-2)^3(-4)(36)=-288\). This is wrong twice: the value of \((x+2)^3\) was misread, and the displayed product actually equals \(-2304\), not \(-288\). The correct substitution is \(q(-3)=(-2)(-1)(-4)(36)\). Since \((-2)(-1)=2\), \(2(-4)=-8\), and \((-8)(36)=-288\), the correct value is \(-288\), which is negative.
This illustrates a useful checking habit: evaluate each factor separately, preserve parentheses and exponents, and then multiply in stages. A sign chart depends on signs, but an incorrect arithmetic calculation can still lead to a wrong sign.
Why Sign Charts Matter
A sign chart compresses a sequence of factor-based checks into a clear record of how a polynomial behaves on the real line. It also provides two independent ways to check the result: direct evaluation at one input per interval and the sign-change rule at each zero. Agreement between those checks makes errors easier to find.
When constructing a chart, distinguish the zero locations from the intervals between them. The polynomial equals zero at its real zeros; the interval signs describe values strictly between those points. A factor with no real zeros does not create a boundary, though its sign still contributes to the product. With those details in place, the chart gives a complete description of the polynomial’s sign on every interval determined by its real zeros.
Check Your Understanding
For each question, use the factors and their signs carefully; when asked for a chart, include every interval determined by the distinct real zeros.
- For \(f(x)=(x-1)^2(x+2)\), list the real zeros, their multiplicities, and the intervals they determine.
- For \(g(x)=-(x+1)(x-2)^4\), state whether the sign changes or stays the same across each zero.
- Choose one test input in each interval and determine the signs of \(h(x)=2(x+3)^3(x-1)^2\).
- Why is one test input sufficient to determine the sign throughout an interval containing no real zero?
- Does \(x^2+9\) create a real-zero boundary in a sign chart? Explain why or why not.
