Polynomial functions: zeros, factors and end behaviour
The factored form tells you where the graph meets the x-axis, whether it crosses there, and which way the two ends go.
Which of the following could define f?
In the xy-plane, the graph of the polynomial function f has exactly two x-intercepts, at (−3, 0) and (5, 0), and f has degree 3.
An x-intercept at (−3, 0) needs the factor (x + 3), because x + 3 = 0 at x = −3; an intercept at (5, 0) needs (x − 5). Three factors in all give degree 3, and repeating (x − 5) supplies the third factor without adding a fourth intercept: f(x) = (x + 3)(x − 5)2. A flips every sign, giving intercepts at 3 and −5, and it is degree 2. C has the right degree but three intercepts — the extra (x + 5) puts one at (−5, 0). D repeats both factors and is degree 4, not 3.
1
“Factor”, “zero” and “x-intercept” are three names for one fact.
(x − r) is a factor of f(x) ⟺ f(r) = 0 ⟺ the graph of y = f(x) meets the x-axis at (r, 0). Any one of the three hands you the other two.
f(x) = (x + 2)(x − 1)(x − 4)
f(1) = (1 + 2)(1 − 1)(1 − 4) = (3)(0)(−3) = 0
one factor is zero, so the whole product is zero
f(−2) = (0)(−3)(−6) = 0 and f(4) = (6)(3)(0) = 0
each factor gets its turn
zeros: −2, 1 and 4
so the graph crosses at (−2, 0), (1, 0) and (4, 0)
Read the arrow the other way too. If a stem says (x − 3) is a factor of p(x), it has told you p(3) = 0 — and that single equation is usually enough to find whatever constant the question is hiding.
p(x) = x³ + 2x² + kx − 20, and (x − 2) is a factor
so p(2) must equal 0
p(2) = 8 + 8 + 2k − 20 = 2k − 4
substitute, do not divide
2k − 4 = 0, so k = 2
one line instead of long division
2
Read the zeros straight off the factors, and mind the sign.
The zero is the number that makes the bracket 0, which is the opposite of the number you see: (x + 7) gives the zero −7.
(x − 6) → zero 6(x + 7) → zero −7(2x − 5) → zero 52(3x + 1) → zero −13
A constant out front scales the graph vertically but cannot move a zero: 3(x − 2)(x + 5), −(x − 2)(x + 5) and 0.4(x − 2)(x + 5) all have zeros 2 and −5. What the constant does control is how tall the graph is, so one extra point pins it down.
f(x) = a(x + 1)(x − 4)2, and the graph passes through (2, −18)
zeros already known; only a is missing
f(2) = a(2 + 1)(2 − 4)2 = a(3)(4) = 12a
square before you multiply: (−2)² = 4
12a = −18, so a = −32
negative, so the graph is flipped over the x-axis
The y-intercept is not a zero
Zeros come from setting y = 0 and solving for x. The y-intercept comes from setting x = 0: for f(x) = (x + 2)(x − 1)(x − 4) it is f(0) = (2)(−1)(−4) = 8, the point (0, 8). Every question that says "intercept" tells you which axis it means — read that word before you compute.
3
A repeated factor touches the axis instead of crossing it.
Multiplicity is how many times a factor appears. Odd multiplicity → the graph crosses. Even multiplicity → the graph touches the axis and turns back the way it came.
h(x) = (x + 1)(x − 3)2
Factor
Multiplicity
At that zero the graph…
(x + 1)
1 (odd)
crosses, going straight through
(x − 3)²
2 (even)
touches and turns back — the sign of h never changes
(x − 3)³
3 (odd)
crosses, but flattens out as it does
The reason is sign. Just left of x = 3, (x − 3) is negative; squared, it is positive. Just right of 3 it is positive; squared, still positive. A factor that never changes sign cannot let the graph change sides.
Multiplicity also counts towards the degree: (x + 1)(x − 3)² is degree 1 + 2 = 3, even though only two different numbers are zeros.
4
Degree and the leading coefficient decide the two ends.
Far out to the left and right, the highest-power term swamps everything else, so those two pieces of information are all the ends depend on.
Degree
Leading coefficient
Left end
Right end
odd
positive
down
up
odd
negative
up
down
even
positive
up
up
even
negative
down
down
Odd degree sends the ends opposite ways, like a line; even degree sends them the same way, like a parabola. A negative leading coefficient flips both.
f(x) = −2(x − 1)(x + 3)(x − 5)
three factors, so degree 3
leading term: (−2)(x)(x)(x) = −2x³
multiply only the leading pieces; ignore the numbers in the brackets
odd degree, negative coefficient → up on the left, down on the right
as x increases without bound, f(x) decreases without bound
Do not expand to find the degree
Multiplying (x − 1)(x + 3)(x − 5) out costs a minute and risks an arithmetic slip. Count factors instead, remembering that a square counts twice and a cube three times, and take the leading coefficient by multiplying the front numbers only.
5
“For how many values of x does f(x) = 0?” means count the distinct zeros.
Not the degree, not the number of factors — the number of different x-values that make the product zero.
f(x) = (x + 3)2(x − 1)(x − 6)
degree 4, but the zeros are −3, 1 and 6 → 3 values
g(x) = (x − 2)3
degree 3, one zero → 1 value
k(x) = (x2 + 4)(x − 5)
x² + 4 = 0 has no real solution → 1 value
A quadratic factor is only worth two zeros if it actually factors over the real numbers. Check its discriminant, or just notice that x² + 4 is never 0 because x² is never negative.
The same count answers "how many x-intercepts does the graph have", because every distinct zero gives exactly one point where the curve meets the x-axis — whether it crosses there or only touches.
6
“Which could be the graph of f?” — check four things in order.
Zeros, then cross or touch at each one, then the two ends, then the y-intercept. The first mismatch eliminates a choice; you rarely need all four.
f(x) = −(x + 2)(x − 3)2
zeros: −2 and 3
any graph with a third intercept is out
crosses at −2, touches at 3
multiplicity 1, then multiplicity 2
degree 3, leading term −x³ → up on the left, down on the right
the minus sign out front is doing all the work
f(0) = −(2)(−3)2 = −(2)(9) = −18
so the curve passes below the origin, through (0, −18)
two intercepts · one crossing, one touch · up-left, down-right · through (0, −18)
The extra factor that quietly adds an intercept
In the opening problem, (x + 3)(x − 5)(x + 5) has the right degree and contains both required factors, which is exactly what makes it tempting. But (x + 5) is a zero of its own, at x = −5, so that graph meets the x-axis three times and the stem said twice. A repeated factor raises the degree without adding an intercept; a new factor raises both.
Desmos is on your screen — use it
Type the factored form straight in, brackets and all: y = −(x + 2)(x − 3)^2. The graph appears with its zeros in place, and clicking a point on the curve reports the coordinates, so you can confirm a touch rather than a crossing by zooming in near x = 3. On a "which could define f" item, graphing two of the four choices is usually faster than reasoning about all four.
✓
Check: does each factor do what the graph shows?
the cubic(x + 2)(x − 1)(x − 4)three crossings, at −2, 1, 4 ✓
the repeat(x + 1)(x − 3)²crosses at −1, touches at 3 ✓
the constant12a = −18a = −1.5, so the graph flips ✓
Remember the flow
Factor ⟺ zero · even multiplicity touches · odd degree splits the ends
Count distinct zeros, not factors, and let the leading coefficient's sign flip the picture.
Now you try
Practice problem
Which of the following could define f?
In the xy-plane, the graph of the polynomial function f has x-intercepts at (−4, 0) and (3, 0) only. The graph crosses the x-axis at (−4, 0) and touches it at (3, 0), and as x increases without bound, f(x) decreases without bound.
Three conditions, three checks. The intercepts −4 and 3 need the factors (x + 4) and (x − 3); crossing at −4 needs (x + 4) to odd power and touching at 3 needs (x − 3) to even power; and "decreases without bound" on the right, with odd degree 3, needs a negative leading coefficient. Only f(x) = −(x + 4)(x − 3)2 does all three. A has every zero right but a positive leading coefficient, so its right end goes up instead of down. B flips the signs inside the brackets, putting the intercepts at 4 and −3. D swaps the multiplicities, so it touches at −4 and crosses at 3 — the mirror image of what the stem describes.
Hint: take the three requirements one at a time and cross off a choice the moment it fails one. The signs inside the brackets settle where; the exponents settle cross or touch; the sign in front settles which way the ends go.
