Introduction
A polynomial can equal zero at a point without changing sign there. At some zeros, its values on the two sides have opposite signs; at others, they have the same sign. The exponent of the factor responsible for the zero tells us which behavior occurs.
This distinction is useful when interpreting polynomial values near a zero. It also prepares us to understand broader sign analysis later. Here we focus on what happens close to one zero at a time, rather than constructing a sign chart over an entire domain.
Multiplicity of a Zero
The Factor Theorem says that if a polynomial \(P\) satisfies \(P(a)=0\), then \(x-a\) is a factor of \(P(x)\). If the factor occurs more than once, its exponent records the multiplicity of the zero.
The condition \(q(a)\ne0\) matters: if \(q(a)=0\), then \(x-a\) is another factor of \(q\), so the exponent was not yet as large as possible. The factor \(q(x)\) is the remaining factor after the full power at \(a\) has been removed.
The multiplicity is even if \(m\) is divisible by \(2\), and odd if \(m\) is not divisible by \(2\). For instance, \((x-a)^2\) and \((x-a)^4\) have even exponents, while \((x-a)^1\) and \((x-a)^3\) have odd exponents.
The Local Sign Rule
On the left of \(a\), the difference \(x-a\) is negative. On the right, it is positive. Raising these values to an even power makes the result positive on both sides; raising them to an odd power preserves the negative sign on the left and the positive sign on the right. The remaining factor \(q(x)\) does not interfere sufficiently close to \(a\), because \(q(a)\ne0\) and a polynomial stays close to its value at \(a\) when \(x\) is close to \(a\).
“Sufficiently close” is important. The theorem describes a neighborhood of \(a\) in which the remaining factor does not vanish or change sign. It does not claim that \(q(x)\) has the same sign everywhere.
Why the Rule Works
Here is a proof that includes the reason the remaining factor keeps its sign. Since \(q\) is a polynomial, it is continuous at \(a\). Because \(q(a)\ne0\), continuity allows us to choose a positive distance \(\delta\) such that whenever \(|x-a|<\delta\), the value \(q(x)\) differs from \(q(a)\) by less than half the nonzero amount \(|q(a)|\).
If \(q(a)>0\), this inequality gives \(q(x)>q(a)-q(a)/2=q(a)/2>0\). If \(q(a)<0\), it gives \(q(x)<q(a)+|q(a)|/2=q(a)/2<0\). Thus \(q(x)\) has the same sign as \(q(a)\) throughout this neighborhood.
Now take \(x\) in this neighborhood, with \(x\ne a\), and use the factorization \(P(x)=(x-a)^m q(x)\). On the left, \(x-a<0\); on the right, \(x-a>0\). For even \(m\), \((x-a)^m>0\) on both sides, so multiplication by \(q(x)\), whose sign is fixed nearby, gives the same sign on both sides. For odd \(m\), \((x-a)^m\) is negative on the left and positive on the right. Multiplication by the same-sign \(q(x)\) therefore gives opposite signs. This proves the rule.
At \(x=a\) itself, \(P(a)=0\), so \(P\) has no positive or negative sign at that point. The comparison is between values immediately to its left and right.
Worked Examples
Worked Example: An Even-Multiplicity Zero
Determine the multiplicity of \(x=2\) for \(P(x)=x^3-3x^2+4\), and decide whether the sign changes across that zero.
First evaluate the polynomial at \(2\):
Thus \(x-2\) is a factor. Dividing the coefficients \(1,-3,0,4\) synthetically by \(2\) gives quotient coefficients \(1,-1,-2\) and remainder \(0\). Therefore, the quotient is \(x^2-x-2\). Factoring it gives
Since the remaining factor \(q(x)=x+1\) satisfies \(q(2)=3\ne0\), the multiplicity of \(2\) is exactly \(2\), which is even. The sign does not change across \(2\), sufficiently close to the zero.
For a direct check, evaluate on each side. At \(x=1.9\), the factorized expression gives
At \(x=2.1\), it gives
Both values are positive, as predicted. The check is consistent with the factorization, since \((x-2)^2(x+1)\) equals the original polynomial.
Worked Example: An Odd-Multiplicity Zero with a Positive Remaining Factor
For \(P(x)=(x+1)^3(2-x)\), determine the sign on each side of \(a=-1\), close to \(a\).
Here, \(x+1=x-(-1)\), so the factor at \(a=-1\) is \((x+1)^3\). The remaining factor is \(q(x)=2-x\), and
Because the remaining factor is nonzero at \(-1\), the zero has multiplicity exactly \(3\), which is odd. The sign must therefore reverse: since \(q(-1)\) is positive, the polynomial is negative just to the left and positive just to the right.
Check the signs at \(x=-1.1\) and \(x=-0.9\). On the left,
On the right,
The values have opposite signs. At the zero itself, \(P(-1)=0^3(3)=0\), as required.
Worked Example: An Even-Multiplicity Zero with a Negative Remaining Factor
For \(P(x)=-2(x-1)^4(x+3)\), determine whether the sign changes across \(x=1\), and identify the nearby signs.
The remaining factor is \(q(x)=-2(x+3)\), and its value at \(1\) is
Since \(q(1)\ne0\), the multiplicity of \(1\) is exactly \(4\). This exponent is even, so the sign stays the same on both sides. Because \(q(1)<0\), the nearby values on both sides are negative.
At \(x=0.9\),
At \(x=1.1\),
Both checks are negative, in agreement with the rule. Also, \(P(1)=0\), because the factor \((1-1)^4\) is zero.
What Multiplicity Does—and Does Not—Tell You
Multiplicity answers whether the sign reverses locally: odd multiplicity means opposite signs on the two sides, while even multiplicity means the same sign. It does not by itself say which side is positive. For that, inspect the remaining factor’s value \(q(a)\). If \(q(a)>0\), an odd-multiplicity zero gives a negative sign on the left and a positive sign on the right; if \(q(a)<0\), those signs are reversed. With even multiplicity, \(q(a)>0\) gives positive signs on both sides, and \(q(a)<0\) gives negative signs on both.
These conclusions apply close enough to the zero that \(q(x)\) keeps the sign of \(q(a)\). Farther away, the remaining factor may have its own zeros, and the sign may change for other reasons. Multiplicity at \(a\) describes the local behavior due to the factor \(x-a\), not every sign change the polynomial might have elsewhere.
A Common Mistake
A common error is to assume that every time a polynomial equals zero, its sign must change. For the polynomial \(P(x)=(x-2)^2(x+1)\), one might reason incorrectly:
That conclusion does not follow. The factor at \(2\) has exponent \(2\), so it is positive on both sides of \(2\). In addition, \(x+1\) is positive near \(2\). For example, \(P(1.9)=0.029>0\) and \(P(2.1)=0.031>0\), so there is no sign reversal. The correction is to find the factor’s multiplicity first: an even exponent means the sign stays the same locally; an odd exponent means it changes.
Why This Matters
A zero records where a polynomial’s output is zero, while its multiplicity gives more information about how the polynomial behaves around that input. This is why two polynomial zeros can behave differently even though both satisfy \(P(a)=0\). Recognizing the exponent lets you predict a local sign change without substituting many values. The sign on a particular side still depends on the remaining factor, and conclusions are local to the zero under consideration.
Check Your Understanding
Use the factor at the specified zero and, when needed, the value of the remaining factor there.
- If \(P(x)=(x-3)^5(x+2)\), what is the multiplicity of \(3\), and does the sign change across it?
- If \(P(x)=-4(x+1)^2(x-5)\), what is the multiplicity of \(-1\), and what is the sign on each side sufficiently close to \(-1\)?
- For \(P(x)=(x-4)^3q(x)\), where \(q(4)<0\), which side of \(4\) has positive values close to \(4\)?
- Why must the remaining factor satisfy \(q(a)\ne0\) for \(m\) to be the multiplicity in \(P(x)=(x-a)^m q(x)\)?
- Does an even-multiplicity zero guarantee positive values on both sides? Explain what else determines the signs.
