Zeros Carry More Information Than Their Locations
A polynomial’s real zeros tell us where its graph meets the horizontal axis. But the location of a zero alone does not tell us whether the graph passes through the axis or turns back to the same side. The multiplicity of the zero provides that information.
For example, \(x=0\) is a zero of both \(f(x)=x\) and \(g(x)=x^2\). The graph of \(f\) crosses the axis there, while the graph of \(g\) touches the axis and remains on the same side. This difference is useful when sketching a polynomial: multiplicities describe local behavior at each zero, while degree and leading coefficient describe the behavior at the far left and right.
Multiplicity of a Polynomial Zero
The factor \(q\) contains the remaining polynomial factors. The condition \(q(a)\ne 0\) ensures that \(x-a\) does not divide \(q\) as well; otherwise the displayed power would not be the full multiplicity.
A zero is odd-multiplicity when its multiplicity is odd, such as \(1\) or \(3\). It is even-multiplicity when its multiplicity is even, such as \(2\) or \(4\). These two cases give different sign behavior on the two sides of the zero.
Why Odd Multiplicity Crosses and Even Multiplicity Touches
Near a zero \(a\), write the polynomial as \(p(x)=(x-a)^m q(x)\), with \(q(a)\ne 0\). Because \(q\) is continuous and \(q(a)\) is nonzero, \(q(x)\) has the same sign as \(q(a)\) for all \(x\) sufficiently close to \(a\). Thus the sign change across \(a\) is determined by \((x-a)^m\).
Here, “crosses” means that the polynomial has opposite signs immediately to the left and right of the zero. “Touches but does not cross” means that its sign is the same on both sides. To see why the theorem holds, compare the signs of the power factor:
For odd \(m\), the factor changes sign at \(a\); multiplying by \(q(x)\), whose sign stays fixed near \(a\), still gives a sign change in \(p(x)\). For even \(m\), the factor is positive on both sides of \(a\); multiplying by a factor with fixed sign leaves the sign of \(p(x)\) the same on both sides. Since \(p(a)=0\), these two cases are precisely crossing and touching, respectively.
This argument describes the behavior close to the zero. It does not say that the graph has no other features farther away. To determine signs over the full real line, combine the behavior at each zero with the signs on the intervals between zeros.
Combining Multiplicities with End Behavior
For a polynomial written in factored form, the real zeros divide the real line into intervals. Within any one such interval, none of the factors corresponding to those zeros changes sign. As a result, the polynomial’s sign is constant on that interval. At each zero, the sign changes for odd multiplicity and stays the same for even multiplicity.
The sign on the far-left interval can be determined from the leading term. If a polynomial has degree \(n\) and leading coefficient \(c\), then its end behavior is determined by \(cx^n\): for sufficiently large \(|x|\), the leading term dominates the lower-degree terms. In particular, on the far-left side, the sign is the sign of \(c(-1)^n\); on the far-right side, it is the sign of \(c\).
Use the factored form to list each distinct zero and the power of its factor.
Use the degree and leading coefficient to establish the signs at the far left and far right.
Flip the sign at an odd-multiplicity zero; keep it unchanged at an even-multiplicity zero.
Mark a crossing at odd multiplicity and a touch without crossing at even multiplicity.
Worked Examples
Worked Example: A Touch Followed by a Crossing
Analyze the zeros, signs, and end behavior of \(f(x)=(x+2)^2(x-1)^3\).
The zero \(x=-2\) has multiplicity \(2\), and \(x=1\) has multiplicity \(3\). The degree is \(2+3=5\), and the leading coefficient is \(1\). Since the degree is odd and the leading coefficient is positive, \(f(x)\) is negative far to the left and positive far to the right.
Starting with a negative sign to the left of \(-2\), the even multiplicity at \(-2\) means the sign stays negative between \(-2\) and \(1\). The odd multiplicity at \(1\) means the sign changes to positive to the right of \(1\). Therefore the graph touches the axis at \(-2\) and crosses it at \(1\).
Check these interval signs by substitution. At \(x=-3\), \(f(-3)=(-1)^2(-4)^3=1(-64)=-64\). At \(x=0\), \(f(0)=2^2(-1)^3=4(-1)=-4\). At \(x=2\), \(f(2)=4^2(1)^3=16\). These values agree with the predicted signs. Also, \(f(-2)=0^2(-3)^3=0\) and \(f(1)=3^2(0)^3=0\), confirming both intercepts.
Worked Example: Negative Leading Coefficient
Analyze \(g(x)=-(x+1)(x-2)^4\).
The zeros are \(-1\), with multiplicity \(1\), and \(2\), with multiplicity \(4\). The degree is \(1+4=5\), and the leading coefficient is \(-1\). An odd-degree polynomial with negative leading coefficient is positive far to the left and negative far to the right.
At \(-1\), the multiplicity is odd, so the sign changes from positive to negative. At \(2\), the multiplicity is even, so the sign remains negative. Thus the graph crosses at \(-1\), then touches the axis at \(2\), staying below it on both sides.
Check the signs with values from each interval. At \(x=-2\), \(g(-2)=-((-1)(-4)^4)=-((-1)(256))=256\), which is positive. At \(x=0\), \(g(0)=-(1)(-2)^4=-16\), which is negative. At \(x=3\), \(g(3)=-(4)(1)^4=-4\), also negative. The roots themselves check directly: \(g(-1)=-(0)(-3)^4=0\), and \(g(2)=-(3)(0)^4=0\). The signs and zeros agree with the multiplicity analysis.
Worked Example: Several Real Zeros and Both Ends
Use multiplicity and end behavior to analyze \(h(x)=(x+1)^3(x-2)^2(x-4)\).
The zeros are \(-1\) with multiplicity \(3\), \(2\) with multiplicity \(2\), and \(4\) with multiplicity \(1\). The degree is \(3+2+1=6\), and the leading coefficient is \(1\). Because the degree is even and the leading coefficient is positive, the polynomial is positive at both ends.
Beginning with a positive sign on the far-left interval, the sign changes at \(-1\) because its multiplicity is odd. It stays negative across \(2\), whose multiplicity is even, and changes to positive across \(4\), whose multiplicity is odd. The graph therefore crosses at \(-1\), touches at \(2\), and crosses at \(4\).
Verify the interval signs by substitution. At \(x=-2\), \(h(-2)=(-1)^3(-4)^2(-6)=(-1)(16)(-6)=96\). At \(x=0\), \(h(0)=1^3(-2)^2(-4)=1(4)(-4)=-16\). At \(x=3\), \(h(3)=4^3(1)^2(-1)=64(1)(-1)=-64\). At \(x=5\), \(h(5)=6^3(3)^2(1)=216(9)=1944\). These signs match the sign chart. Substitution also gives \(h(-1)=0\), \(h(2)=0\), and \(h(4)=0\), as required.
A Common Mistake: Treating Every Zero as a Crossing
A frequent error is to mark a crossing at every zero, ignoring its multiplicity. For \(r(x)=(x-3)^2\), this incorrect reasoning might say: “The graph is zero at \(x=3\), so it crosses the axis there; the sign changes from negative to positive.” But the sign does not change. For every \(x\ne 3\), \((x-3)^2>0\), so \(r(x)\) is positive on both sides of \(3\). At the zero, \(r(3)=0\). The graph touches the axis at \(x=3\) and stays above it.
Why This Information Matters in a Sketch
End behavior gives a reliable starting point for the graph’s far-left and far-right directions. Multiplicities then tell how the graph moves from one interval to the next at each real zero. Together, these facts prevent an important kind of sketching error: drawing the correct intercepts but connecting them with signs or crossings that are impossible for the polynomial.
Multiplicity does not determine every detail of a graph between its zeros. Other features may affect its shape there. But for the question of crossing versus touching, and for how the sign changes along the real line, multiplicity gives the needed information.
Check Your Understanding
Use multiplicities and end behavior to answer these questions.
- At a zero with multiplicity \(5\), does the polynomial change sign across the zero?
- What does an even multiplicity tell you about the polynomial’s signs immediately to the left and right of that zero?
- For \(p(x)=-(x-1)^2(x+3)\), identify the multiplicity of each zero and state whether the graph crosses or touches at each one.
- For \(q(x)=2(x+1)^3(x-2)^2\), determine the signs at the far left and far right.
- In a sign chart, what should you do to the sign when passing an odd-multiplicity zero? What about an even-multiplicity zero?
