Zeros and multiplicity
The polynomials (x - 1)(x + 3) and (x - 1)^2(x + 3) have exactly the same zeros. Why does one graph cross the axis at 1 while the other only touches it there?
Where you are 6 in Polynomial functions
Why this exists
The Solving quadratics by factoring turns a factored polynomial into its zeros. Both (x - 1)(x + 3) and (x - 1)^2(x + 3) are zero at x = 1 and x = -3 and nowhere else, so their lists of x-intercepts are the same. Their graphs are not. The first passes through the axis at 1; the second comes down, touches the axis at 1, and goes back up.
The list has lost how many times each factor appears. Keeping that count, the multiplicity, tells you what the graph does at each zero without plotting a point.
Definition
Let p be a polynomial and c a real zero of p, so that x - c is a factor of p by the The remainder and factor theorems. The multiplicity of c is the number of times the factor x - c appears: the largest positive integer k for which (x - c)^k is a factor of p. Equivalently,
The condition q(c) \neq 0 says that no copy of x - c is left inside q. A zero of multiplicity 1 is a simple zero, of multiplicity 2 a double zero, of multiplicity 3 a triple zero.
In (x - 1)^2(x + 3), the zero 1 has multiplicity 2 and -3 has multiplicity 1. A quadratic whose discriminant is zero has a single double zero: x^2 + 10x + 25 = (x + 5)^2 has one at -5, and its parabola touches the axis there, at its vertex.
Intuition
Near a zero c, the factor (x - c)^k is tiny and changes fast, while q(x) hardly moves from q(c). So near c the graph looks like y = (x - c)^k stretched by q(c): everything comes down to y = x^k near the origin.
| k | x^k at x = -0.1 | x^k at x = 0.1 |
|---|---|---|
| 1 | -0.1 | 0.1 |
| 2 | 0.01 | 0.01 |
| 3 | -0.001 | 0.001 |
| 4 | 0.0001 | 0.0001 |
Two things show. The sign: an odd power is negative on the left and positive on the right, so it crosses; an even power is positive on both sides, so it touches and turns back. The size: each extra power makes the output ten times smaller at a distance of 0.1, so a higher power hugs the axis longer before it leaves. That is the flattening.
Results
Crossing or touching. Let c be a real zero of multiplicity k of a polynomial p, with p(x) = (x - c)^k q(x) and q(c) \neq 0.
- If k is odd, the graph crosses the x-axis at c: p has opposite signs just left and just right of c.
- If k is even, the graph touches the axis at c and turns back: p has the same sign on both sides, and (c, 0) is a turning point (Turning points, maximums and minimums).
- If k = 1 the graph crosses at a slant, like a line. If k \geq 2 it meets the axis level; for odd k \geq 3 it flattens as it crosses, like y = x^3. The larger k, the flatter.
- Near c, p(x) \approx q(c)\,(x - c)^k. For even k the graph is above the axis when q(c) > 0 and below when q(c) < 0; for odd k it crosses going up when q(c) > 0 and going down when q(c) < 0.
Counting. If the distinct real zeros of a polynomial of degree n have multiplicities k_1, k_2, \dots, k_m, then
Equality holds exactly when p is a constant times a product of linear factors with real zeros.
Why it is true
The sign. The polynomial q has finitely many zeros, none at c, and its graph has no breaks (Continuous graphs, jumps and holes), so on some interval around c it keeps the sign of q(c). On that interval the sign of p is the sign of q(c) times the sign of (x - c)^k. To the right of c, x - c > 0, so (x - c)^k > 0. To the left, x - c < 0, and a product of k negative numbers is negative when k is odd and positive when k is even. So for odd k the sign of p switches at c, and for even k it does not.
The flatness. For x \neq c, the slope of the line from (c, 0) to (x, p(x)) is \displaystyle \frac{p(x)}{x - c} = (x - c)^{k - 1} q(x). As x closes in on c this closes in on q(c) \neq 0 when k = 1, a slanted line, and on 0 when k \geq 2, a level one.
The count. Write p(x) = (x - c_1)^{k_1} q_1(x). Since x - c_2 shares no factor with x - c_1, all k_2 copies of x - c_2 in p must sit inside q_1, the way every factor 3 of 2^5 \cdot m must sit inside m. Taking out each zero in turn gives p(x) = (x - c_1)^{k_1} \cdots (x - c_m)^{k_m}\, r(x) for a polynomial r. Degrees add when polynomials multiply, so k_1 + \dots + k_m + \deg r = n, and the sum is at most n, with equality exactly when r is a constant.
Worked examples
A double zero and a simple zero
Find the zeros of f(x) = (x - 1)^2(x + 3) with their multiplicities, and say what the graph does at each.
SolutionThe zeros are 1, of multiplicity 2, and -3, of multiplicity 1, adding to 3, the degree.
At 1 the multiplicity is even, so the graph touches and turns back. Near 1 the other factor x + 3 is close to 4, so f(x) \approx 4(x - 1)^2, which is never negative: the graph comes down to the axis from above and goes back up.
At -3 the multiplicity is odd, so the graph crosses. Near -3 the factor (x - 1)^2 is close to 16, so f(x) \approx 16(x + 3), negative on the left and positive on the right: it crosses going up, at a slant.
Check. f(-4) = (-5)^2(-1) = -25 and f(-2) = (-3)^2(1) = 9 change sign across -3, while f(0) = 3 and f(2) = 5 are both positive either side of 1. Flatness shows too: a step of 0.1 from each zero gives f(1.1) = 0.041 but f(-2.9) = 1.521.
A triple zero at the origin
Describe the graph of g(x) = x^3(x - 2) at each zero.
SolutionSince x^3 = (x - 0)^3, the zero 0 has multiplicity 3, and 2 has multiplicity 1. The total is 4, the degree.
At 0 the multiplicity is odd and at least 3: the graph crosses, flattening as it does. Near 0 the factor x - 2 is about -2, so g(x) \approx -2x^3, positive just left of 0 and negative just right: it crosses going down. How flat: g(0.1) = (0.001)(-1.9) = -0.0019.
At 2 the zero is simple. Near 2, x^3 \approx 8, so g(x) \approx 8(x - 2): the graph crosses going up, at a slant.
Check. g(-1) = (-1)(-3) = 3, g(1) = (1)(-1) = -1 and g(3) = 27: positive, negative, positive, matching both crossings. Both ends rise, as they must for even degree and a positive leading coefficient (Degree, end behavior and the shape of a polynomial graph).
A double and a triple zero
Describe the graph of h(x) = (x + 2)^2(x - 1)^3 near each zero, and find where it lies above and below the x-axis.
SolutionThe zeros are -2, of multiplicity 2, and 1, of multiplicity 3; the degree is 5. Track the sign of each factor between the zeros.
| x < -2 | -2 < x < 1 | x > 1 | |
|---|---|---|---|
| (x + 2)^2 | + | + | + |
| (x - 1)^3 | - | - | + |
| h(x) | - | - | + |
A square is never negative, so the factor (x + 2)^2 cannot change the sign: the graph is below the axis on both sides of -2. It rises to touch the axis at (-2, 0) and turns back down, a peak. Near -2, (x - 1)^3 \approx -27, so h(x) \approx -27(x + 2)^2.
The factor (x - 1)^3 changes sign at 1, so the graph crosses there, flattened, going up: h(x) \approx 9(x - 1)^3.
So h(x) < 0 for x < 1 except at x = -2, where h = 0, and h(x) > 0 for x > 1.
Check. h(-3) = (1)(-64) = -64, h(0) = (4)(-1) = -4 and h(2) = (16)(1) = 16.
Where is it positive?
For which x is s(x) = -(x - 4)(x + 1)^2 positive?
SolutionThe zeros are 4, of multiplicity 1, and -1, of multiplicity 2. The graph has no breaks, so its sign can change only at a zero.
Near -1, the factor -(x - 4) is about -(-5) = 5, so s(x) \approx 5(x + 1)^2: the graph touches from above and is positive on both sides of -1.
Near 4, (x + 1)^2 \approx 25, so s(x) \approx -25(x - 4): positive just left of 4 and negative just right. It crosses going down.
So s(x) > 0 for every x < 4 except x = -1, where s(-1) = 0; and s(x) < 0 for x > 4.
Check. s(-2) = -(-6)(1) = 6, s(0) = -(-4)(1) = 4 and s(5) = -(1)(36) = -36. The leading term is -x^3, so the left end rises and the right end falls, as found.
Reading multiplicities off a graph
The graph of a polynomial p of degree 6 is shown whole and in a close-up near x = 1 and x = 2. Find the multiplicity of each zero and write p(x) up to a constant factor.
The whole graph crosses at -3 at a slant. Between 1 and 2 it seems to lie on the axis: its outputs there are smaller than 0.2 in size, which a scale of hundreds cannot show. In the close-up, the graph touches at 1 and turns back down, and crosses at 2 lying flat.
So -3 has odd multiplicity, and since the crossing is at a slant, multiplicity 1. The zero 1 has even multiplicity, at least 2. The zero 2 has odd multiplicity and is flat, so at least 3. These already add to 1 + 2 + 3 = 6, the degree, and the total cannot be more. So the multiplicities are exactly 1, 2 and 3, and p has no other zeros:
for a constant a. Both ends rise, so a > 0. One known point fixes a (Building a polynomial from its zeros); this graph was drawn with a = 1, for which p(0) = 3 \cdot 1 \cdot (-8) = -24.
A rollercoaster that touches the ground
Dagrun designs a stretch of rollercoaster track whose height in meters above the ground, x meters along, is
for 0 \leq x \leq 44. A negative height means the track runs in a tunnel. (a) Where is the track at ground level, and where does it go into or come out of the tunnel? (b) How high is it at x = 0, x = 20 and x = 35? (c) Edvard says the track meets the ground at the same steepness at all three points. Is he right?
Solution(a) H(x) = 0 at x = 10, a double zero, and at x = 30 and x = 40, both simple. At 10 the multiplicity is even, and the rest of the formula there is \displaystyle \frac{(-20)(-30)}{2000} = 0.3 > 0: the track comes down, touches the ground and climbs again. At 30 and 40 the multiplicity is odd, so the track passes through ground level. Near 30, \displaystyle H(x) \approx \frac{(20)^2(-10)}{2000}(x - 30) = -2(x - 30), so it goes down into the tunnel; near 40, \displaystyle H(x) \approx \frac{(30)^2(10)}{2000}(x - 40) = 4.5(x - 40), so it comes back up. The tunnel runs from x = 30 to x = 40.
(b) \displaystyle H(0) = \frac{100(-30)(-40)}{2000} = 60 meters, \displaystyle H(20) = \frac{100(-10)(-20)}{2000} = 10 meters, and \displaystyle H(35) = \frac{625(5)(-5)}{2000} = -7.8125: about 7.8 meters below ground, in the tunnel.
(c) No. Near the double zero, H(x) \approx 0.3(x - 10)^2: level as it touches, only 0.003 meters up at 0.1 meters away. At the simple zeros it meets the ground at a slant, dropping about 2 meters per meter at 30 and climbing about 4.5 at 40.
Boundaries
Real zeros only. A complex zero has a multiplicity but no point on the x-axis. A factor such as x^2 + 4 adds to the degree but not to the intercepts, which is why the count is “at most n”; counting complex zeros too, it is exactly n, the The fundamental theorem of algebra.
Factor completely first. (x^2 - 9)(x + 3) hides a repeated factor: it is (x - 3)(x + 3)^2. A factor (2x - 1)^2 is 4\left(x - \tfrac{1}{2}\right)^2, a double zero at \tfrac{1}{2}. A constant factor is never a zero.
The rule is local. Multiplicity says what the graph does right at a zero, not how high it climbs between zeros. Putting the zeros, the ends and the intercept together is Graphing polynomial functions.
A picture shows the parity, not the number. A touch could be multiplicity 2 or 4, a flat crossing 3 or 5; the degree count may settle it (Example 5). A coarse window can hide a touch, and a simple zero beside a turning point can look flat. The formula decides.
The zero polynomial is zero everywhere and has no multiplicities.
Where this goes next
Graphing polynomial functions sketches a whole graph from factored form, adding the end behavior and the y-intercept to the behavior at each zero. Building a polynomial from its zeros runs the other way, choosing each power to make the graph cross or touch. When no factored form is given, the The rational root theorem supplies candidate zeros. The odd and even rule returns in Graphing rational functions, where a power (x - c)^k in a denominator decides whether the graph leaves a vertical asymptote the same way on both sides.
Exercises
Multiplicity from factored form
Give each zero’s multiplicity and say whether the graph crosses or touches there.
- f(x) = (x + 5)(x - 2)^2
- g(x) = x^2(x + 4)^3
- h(x) = 3(x - 6)^4(x + 1)
- p(x) = -2x(x - 3)^5
- q(x) = (2x - 1)^2(x + 3)
Factor first
Factor completely, then give each real zero’s multiplicity and say whether the graph crosses or touches there.
- x^3 - 6x^2 + 9x
- x^4 - 8x^3 + 16x^2
- x^5 - 4x^3
- (x^2 - 4)(x - 2)^2
- (x^2 + 4)(x - 1)^3
Signs near a zero
- For f(x) = (x - 3)^2(x + 2), is the graph above or below the axis just left and just right of 3? Confirm with f(2.9) and f(3.1).
- Describe the graph of g(x) = -(x + 1)^3(x - 4)^2 near each zero: crossing or touching, and from which side.
- For which x is h(x) = (x - 2)^2(x + 1)(x - 5) negative?
Reading graphs
For each graph, give each zero’s least possible multiplicity, the least possible degree, and the sign of the leading coefficient.
- Graph A
- Graph B
- Graph C
In context
- Grzegorz designs a track whose height in meters is \displaystyle H(x) = \frac{1}{1000}(x - 20)^2(60 - x) at x meters along, for 0 \leq x \leq 65, ending at an underground platform. (a) Where does it touch the ground and climb again, and where does it enter the tunnel? (b) Find its height at x = 0, x = 40 and x = 65.
- Halldor sketches a track that touches the ground and climbs again at three places and passes into a tunnel at one more. What is the least degree of a polynomial that could model it?
Explaining
- Explain why the graph of a polynomial of degree 4 cannot touch the x-axis at one point and cross it at three others.
- Explain why a polynomial whose real zeros all have even multiplicity never changes sign.
- Faustyna says (x - 3)^2 and (x - 3)^4 do the same thing at 3. Compare their values at x = 3.1, say how the graphs differ near 3, and find where the two graphs meet.
Answers
- -5 has multiplicity 1: crosses. 2 has multiplicity 2: touches.
- 0 has multiplicity 2: touches. -4 has multiplicity 3: crosses, flattening.
- 6 has multiplicity 4: touches, flatter than a double zero. -1 has multiplicity 1: crosses. The factor 3 gives no zero.
- 0 has multiplicity 1: crosses at a slant. 3 has multiplicity 5: crosses, flattening.
- (2x - 1)^2 = 4\left(x - \tfrac{1}{2}\right)^2, so \tfrac{1}{2} has multiplicity 2: touches. -3 has multiplicity 1: crosses.
- x(x^2 - 6x + 9) = x(x - 3)^2. 0 has multiplicity 1: crosses. 3 has multiplicity 2: touches.
- x^2(x^2 - 8x + 16) = x^2(x - 4)^2. 0 and 4 each have multiplicity 2: the graph touches at both and is never below the axis.
- x^3(x^2 - 4) = x^3(x - 2)(x + 2). 0 has multiplicity 3: crosses, flattening. 2 and -2 have multiplicity 1: crosses.
- (x - 2)(x + 2)(x - 2)^2 = (x - 2)^3(x + 2). 2 has multiplicity 3: crosses, flattening. -2 has multiplicity 1: crosses.
- x^2 + 4 \geq 4 has no real zeros. 1 has multiplicity 3: crosses, flattening. The real multiplicities total 3, less than the degree 5; the other two zeros, \pm 2i, are complex.
- Above on both sides: near 3, f(x) \approx 5(x - 3)^2. f(2.9) = 0.049 and f(3.1) = 0.051. The graph touches from above.
- At -1, multiplicity 3: g(x) \approx -25(x + 1)^3, so it crosses going down, flattening. At 4, multiplicity 2: g(x) \approx -125(x - 4)^2, so it touches from below. Values: g(-2) = 36, g(0) = -16, g(5) = -216.
- -1 < x < 5 except x = 2. For x > 5 every factor is positive. The graph crosses at 5, touches from below at 2, and crosses back above at -1.
- Crosses at -1 at a slant: multiplicity 1. Touches at 2: even, least 2. Least degree 3. The left end rises and the right end falls: odd degree, negative leading coefficient.
- Crosses at -1 lying flat: odd, least 3. Crosses at 3 at a slant: multiplicity 1. Least degree 4. Both ends rise: even degree, positive leading coefficient.
- Crosses at -2 at a slant: multiplicity 1. Touches at 0 and at 2: even, least 2 each. Least degree 5. The left end rises and the right end falls: odd degree, negative leading coefficient.
- (a) 60 - x = -(x - 60), so 20 is a double zero and 60 a simple one. The track touches the ground at x = 20 and climbs again, since H(x) \approx 0.04(x - 20)^2 there; it passes into the tunnel at x = 60, where H(x) \approx -1.6(x - 60). (b) H(0) = 24 meters, H(40) = 8 meters and H(65) = -10.125: the platform is 10.125 meters below ground.
- 7. Each touch needs multiplicity at least 2 and the crossing at least 1: 2 + 2 + 2 + 1 = 7, at most the degree. Degree 7 is enough: -(x - 1)^2(x - 2)^2(x - 3)^2(x - 4) touches at 1, 2 and 3 and dives below at 4.
- A touch needs even multiplicity, at least 2, and each crossing at least 1, so the multiplicities would add to at least 2 + 1 + 1 + 1 = 5. The total cannot exceed the degree, 4.
- At each zero the sign is the same on both sides, since the multiplicity is even. Between zeros the graph has no breaks and no zeros, so the sign cannot change there either. The graph stays on one side of the axis, touching it at each zero, as x^2(x - 3)^2 does.
- Both touch the axis at 3 from above, since both multiplicities are even. At 3.1 they are 0.01 and 0.0001: the fourth power hugs the axis much more closely, a flatter touch. They meet where (x - 3)^2 = (x - 3)^4, that is (x - 3)^2\left(1 - (x - 3)^2\right) = 0: at x = 2, 3 and 4.