Handles the middle of a polynomial graph. Finds roots by factoring, and uses each root's multiplicity to decide whether the curve crosses the axis, bounces off it, or flattens through. Combines roots, multiplicities and end behaviour into a complete sketch, recovers a formula from a graph, and uses sign changes to guarantee roots exist.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 3 — Polynomial and Rational Functions
§3.4 Graphs of Polynomial Functions, pp. 335-361
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 335-361 — the pages these objectives are drawn from
Warm-up
Meeting the axis and crossing it are not the same thing, and the difference is this lesson's subject.
Discussion prompt
Where does the squaring rule meet the horizontal axis, and does it cross there? What about the cubing rule?
Hint: Look at the sign of the output just before and just after each meeting point.
Answer:
The squaring rule meets the axis at the origin and does not cross: outputs are positive on both sides, so the curve comes down, touches, and goes back up.
The cubing rule meets the axis at the origin and does cross: outputs are negative on the left and positive on the right, so the sign changes as it passes through.
Both roots are at the same place, and both come from the same factor — but one is squared and one is cubed. The exponent decides, and that exponent is what this section calls the multiplicity.
Concept
In factored form, each root's exponent is its multiplicity. An odd multiplicity makes the graph cross the axis; an even one makes it bounce off. Higher multiplicities flatten the approach.
multiplicity — The exponent of a factor in a polynomial's factored form. A root of multiplicity one crosses the axis; an even multiplicity means the graph touches and turns back; an odd multiplicity above one crosses with a flattening.
\[ f(x)=a(x-r_1)^{m_1}(x-r_2)^{m_2}\cdots \]
The reason is a sign argument. Near a root, the factor for that root is the only one changing sign, and all the others keep whatever sign they have. An odd power of a quantity changing sign changes sign too; an even power does not. So the product's sign flips or does not, exactly as the parity says.
Figure (svg): Three close-up views of a graph meeting the horizontal axis: crossing straight through for multiplicity one, bouncing off for multiplicity two, and flattening as it passes through for multiplicity three
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 335-342
Section
Section 1
Concept
The roots of a polynomial are the inputs where it equals zero. In factored form they can be read off directly, since a product is zero exactly when one of its factors is.
The third bullet is worth guarding against. Factoring an x out of every term is the first move, and the root at zero it produces is real and is frequently forgotten because it does not come from a bracket that looks like the others.
Figure (svg): A polynomial in factored form sketched from its roots and end behaviour, with each root labelled by its multiplicity and the corresponding crossing or bouncing behaviour marked
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 335-341
Picture it
Three roots, three behaviours decided by the exponents in the factored form.
Figure (svg): A polynomial in factored form sketched from its roots and end behaviour, with each root labelled by its multiplicity and the corresponding crossing or bouncing behaviour marked
The factored form shows all three roots at a glance. Expanding this polynomial would produce a quartic in which none of them is visible.
Worked example
The root it produces is the one most often missed.
\[ \text{Find the roots of } f(x)=x^3-4x. \]
Factor out the common x
Why: Every term has one.
\[ x(x ^{2} - 4) \]
Factor the remaining difference of squares
Why: Two squares subtracted.
\[ x(x - 2) (x + 2) \]
Set each factor to zero
Why: Three factors, three roots.
\[ x = 0, 2, -2 \]
Check the count against the degree
Why: Degree 3, three roots.
Figure (svg): The solution to Worked example factor out the common factor first shown as a ladder of expressions, one row per legal move
\[ x = 0, \; 2, \; -2 \]
Verify: confirm the root at zero
Why: Substituting zero into the original gives zero minus zero, which is zero — so it genuinely is a root. It came from the common factor rather than from a bracket of the usual shape, which is exactly why it is the one that gets forgotten. Counting the roots against the degree is the habit that catches the omission.
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 336-338
Sorting
Match each polynomial to the first move.
Sort into buckets
Sort each by the technique that starts it.
Worked example
Four terms with no common factor often group in pairs.
\[ \text{Find the roots of } f(x)=x^3+2x^2-9x-18. \]
Group in pairs
Why: First two, then last two.
\[ (x ^{3} + 2 x ^{2}) + (-9 x - 18) \]
Factor each pair
Why: Watch the sign on the second.
\[ x ^{2}(x + 2) - 9(x + 2) \]
Factor out the common bracket
Why: Both pairs share it.
\[ (x + 2) (x ^{2} - 9) \]
Factor the difference of squares
Why: And read all three roots.
\[ (x + 2) (x - 3) (x + 3) \]
Figure (svg): The solution to Worked example factor by grouping shown as a ladder of expressions, one row per legal move
\[ x = -2, \; 3, \; -3 \]
Verify: check one root by substitution
Why: At x equal to 3: twenty seven plus eighteen minus twenty seven minus eighteen, which is zero. The grouping worked. The sign on the second pair is the step to watch: factoring out negative 9 rather than 9 is what makes both brackets match, and factoring out positive 9 would have left brackets that differ by a sign.
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 339-341
Trap
\[ x^3-4x=0 \;\Longrightarrow\; x^2-4=0 \;\Longrightarrow\; x = \pm 2 \]
Divide both sides by x to simplify
Why: The equation becomes a quadratic, which is easier to solve.
Two roots are reported, and the answer omits the third.
Dividing by x throws away the root at zero. The division is only valid when x is not zero, and x equal to zero is precisely one of the solutions.
Factoring instead of dividing keeps it: x times the quantity x squared minus 4 equals zero gives three roots, including the one at zero.
Never divide an equation by an expression containing the variable. Factor it out and set it to zero along with everything else — the two operations look similar and one of them silently loses solutions.
Faded example
Factor the polynomial 3x cubed minus 12x.
Fill in the blanks
3x(x^2-4) = 3x(x-2)(x+2)
Why: Factoring out 3x leaves x squared minus 4, which is a difference of squares and factors into two brackets. The three roots are 0, 2 and negative 2 — and the one at zero comes from the 3x, which is the factor that produces the most frequently overlooked root.
Prediction
A cubic factors into a linear factor and a quadratic with a negative discriminant.
Predict first
How many real roots does it have?
Correct: One, from the linear factor only.
Why: The linear factor contributes one real root and the quadratic contributes none, since its discriminant is negative. So there is exactly one real root — which is consistent with §3.3's guarantee that an odd degree has at least one. The quadratic still contributes two complex roots, bringing the total to three in the complex numbers.
Step zero
You are given a polynomial and asked to find all its real roots.
Discussion prompt
What do you check before attempting any factoring technique, and why?
Hint: Is there something every term has in common?
Answer:
Look for a common factor across every term, including a power of x. Removing it reduces the degree of what remains, which makes every subsequent step easier.
It also immediately produces a root: if an x factors out, then zero is a root. Recording it at that moment is the reliable way not to lose it later, since it comes from a factor that does not look like the others.
Only then should the remaining factor be examined for a special form, a grouping, or the quadratic formula. Attempting those first on the unsimplified polynomial usually means doing harder algebra than necessary.
Section
Section 2
Concept
A factor's exponent is its root's multiplicity. Odd multiplicity means the graph crosses the axis there; even multiplicity means it touches and turns back.
The sign argument is the reason. Near a root only that root's factor changes sign; every other factor keeps whatever sign it has. Raising a sign-changing quantity to an odd power keeps the sign change, and to an even power destroys it — so the product's sign flips exactly when the multiplicity is odd.
Figure (svg): Three close-up views of a graph meeting the horizontal axis: crossing straight through for multiplicity one, bouncing off for multiplicity two, and flattening as it passes through for multiplicity three
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 342-348
Picture it
Each panel shows the graph near a root, with the multiplicity labelled.
Figure (svg): Three close-up views of a graph meeting the horizontal axis: crossing straight through for multiplicity one, bouncing off for multiplicity two, and flattening as it passes through for multiplicity three
Multiplicity one crosses at an angle; two bounces; three crosses with a visible flattening. Higher multiplicities look progressively flatter without changing whether they cross.
Worked example
Each exponent is a multiplicity, and each decides one behaviour.
\[ \text{For } f(x)=(x+1)(x-2)^2(x-5)^3, \text{ describe the behaviour at each root.} \]
Read the first factor
Why: Exponent 1, which is odd.
Read the second
Why: Exponent 2, which is even.
\[ \sqrt{2}:\text{ bounces} \]
Read the third
Why: Exponent 3, odd but greater than one.
Check the degree
Why: Multiplicities sum to 1 plus 2 plus 3.
\[ ^\circ 6 \]
Figure (svg): Three close-up views of a graph meeting the horizontal axis: crossing straight through for multiplicity one, bouncing off for multiplicity two, and flattening as it passes through for multiplicity three
\[ -1 \text{ crosses}, \; 2 \text{ bounces}, \; 5 \text{ crosses flattened} \]
Verify: check the end behaviour is consistent
Why: The degree is 6, even, and the leading coefficient is positive, so both ends go up. Starting from the top left, the curve crosses down at negative 1, bounces off at 2 without crossing, and crosses up at 5 to finish at the top right — which is a consistent story. If it were not, one multiplicity would be wrong.
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 343-345
Sorting
Parity of the multiplicity decides.
Sort into buckets
Sort each root by its behaviour.
Worked example
The behaviour at each root reports its multiplicity's parity.
\[ \text{A degree } 4 \text{ graph crosses at } -2, \text{ bounces at } 1, \text{ and crosses at } 3. \]
Read the first behaviour
Why: Crossing means odd, and the smallest odd is 1.
\[ -2:\text{ multiplicity } 1 \]
Read the second
Why: Bouncing means even, smallest is 2.
\[ 1:\text{ multiplicity } 2 \]
Read the third
Why: Crossing again.
\[ 3:\text{ multiplicity } 1 \]
Check the sum
Why: One plus two plus one is four.
\[ \text{sums to } ^\circ 4 \]
Figure (svg): The solution to Worked example find multiplicities from a graph shown as a ladder of expressions, one row per legal move
\[ (x+2)(x-1)^2(x-3) \]
Verify: confirm the sum check did real work
Why: Had the graph been degree 6 with the same three behaviours, the multiplicities could not all have been the minimum, and the extra two would have to be distributed — a flattened crossing would indicate a 3 rather than a 1. The sum check is what turns a parity reading into a definite answer.
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 346-348
Trap
\[ f(x)=(x-3)^2(x+1) \;\Longrightarrow\; \text{degree } 2 \]
Count the brackets
Why: There are two of them, so the degree is taken as 2.
The polynomial is described as a quadratic.
The exponents count too. The squared bracket contributes 2 to the degree and the other contributes 1, so the degree is 3.
Expanding confirms it: the highest term is x cubed. A cubic has different end behaviour from a quadratic, so this error changes the shape of the whole sketch.
Add the multiplicities to get the degree, not the number of distinct factors. The two agree only when every multiplicity is one.
Prediction
A polynomial in factored form has a linear factor, a squared factor and a cubed factor.
Predict first
What is its degree?
Correct: 6, the sum of the multiplicities.
Why: Each factor contributes its own exponent to the degree, so they add: 1 plus 2 plus 3 gives 6. Counting the brackets gives 3, which is the number of distinct roots rather than the degree, and taking the largest exponent gives 3 as well — both are wrong for the same reason.
Faded example
A degree 6 polynomial crosses at 0, bounces at 2, and crosses with flattening at negative 1.
Fill in the blanks
f(x) = a\,x(x-2)^2}(x+1)^3}
Why: A plain crossing at zero gives multiplicity 1, a bounce at 2 gives the smallest even value 2, and a flattened crossing at negative 1 gives the smallest odd value above one, which is 3. Those sum to 6, matching the stated degree — which is the check that confirms each multiplicity was taken at its smallest consistent value rather than guessed.
Socratic
The crossing rule is usually memorised as odd crosses, even bounces.
Discussion prompt
Explain why, using the sign of each factor near a root.
Hint: Which factors change sign as x passes through one particular root?
Answer:
Near one root, only that root's factor changes sign. Every other factor is evaluated at inputs far from its own root, so it keeps whatever sign it has throughout the neighbourhood.
So the product's sign change depends entirely on that one factor raised to its multiplicity. A sign-changing quantity raised to an odd power still changes sign; raised to an even power it does not, since an even power is never negative.
The graph crosses exactly when the output changes sign, so it crosses exactly when the multiplicity is odd. That is the whole argument, and it explains the flattening too: a higher power of a small number is smaller still, so the curve hugs the axis longer on the approach.
Section
Section 3
Concept
A complete sketch needs only the roots with their multiplicities, the end behaviour, and the y-intercept. Everything else follows.
The order matters. Drawing the ends first means the curve is anchored at both extremes, and the roots then constrain what happens in between, so the sketch has only one consistent way to be drawn. Starting in the middle and working outward usually produces a curve whose ends are wrong.
Figure (svg): A polynomial in factored form sketched from its roots and end behaviour, with each root labelled by its multiplicity and the corresponding crossing or bouncing behaviour marked
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 348-354
Picture it
Three roots, one of them a bounce, and both ends up.
Figure (svg): A polynomial in factored form sketched from its roots and end behaviour, with each root labelled by its multiplicity and the corresponding crossing or bouncing behaviour marked
The bounce at 1 is what forces the shallow dip between 1 and 3 rather than a straight run. Change that factor's exponent to 1 and the whole middle of the picture changes.
Worked example
Ends first, then the roots.
\[ \text{Sketch } f(x)=(x+2)(x-1)^2(x-3). \]
Find the degree and leading coefficient
Why: Multiplicities sum to 4; leading coefficient positive.
\[ ^\circ 4, a > 0 \]
Draw the end behaviour
Why: Even degree, positive coefficient.
Mark the roots and behaviours
Why: Reading the exponents.
Find the y-intercept
Why: Substitute zero.
\[ (2) (1) (-3) = -6 \]
Figure (svg): A polynomial in factored form sketched from its roots and end behaviour, with each root labelled by its multiplicity and the corresponding crossing or bouncing behaviour marked
\[ \text{ends up}; \; -2 \text{ cross}, \; 1 \text{ bounce}, \; 3 \text{ cross}; \; (0,-6) \]
Verify: check the story is consistent
Why: Coming down from the top left, the curve crosses at negative 2 into negative territory, passes the y-intercept at negative 6, comes up to touch zero at 1 and bounces back down, then crosses up at 3 to finish at the top right. Every feature agrees, and the negative y-intercept confirms the curve is below the axis between negative 2 and 1.
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 349-351
Ranking
The order that makes the sketch come out right.
Put in order
Why: Drawing the ends first anchors the curve at both extremes, which leaves only one consistent way to join up. Marking the roots and the y-intercept adds the constraints in between, and the joining is done last because it is the step that has to respect all the others.
Worked example
The sketch must be consistent with the degree's bound.
\[ \text{How many turns does the sketch above have, and is that allowed?} \]
Trace from the left
Why: Down from the top, turning at a minimum.
Continue to the bounce
Why: Up to touch the axis at 1, which is a turn.
Continue past it
Why: Down to a minimum, then up.
Compare with the bound
Why: Degree 4 allows at most 3.
Figure (svg): The solution to Worked example count the turning points shown as a ladder of expressions, one row per legal move
\[ 3 \text{ turns, and the bound for degree } 4 \text{ is } 3 \]
Verify: notice the bounce is itself a turn
Why: A bounce is a turning point by definition: the curve reverses direction there. That is worth counting, because it is easy to look for turns only between roots and miss the one sitting on the axis. Here the three turns are the minimum on the left, the bounce, and the minimum on the right.
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 352-354
Error analysis
A student sketches a polynomial with a squared factor.
Annotate
On: \( f(x)=(x-1)^2(x-4) \;\Longrightarrow\; \text{crosses at } 1 \text{ and at } 4 \)
Check the exponent on every factor before deciding what happens at its root. A sketch that crosses where it should bounce has the wrong shape over an entire interval, not just at a point.
Prediction
A polynomial crosses at 2 and has no other roots between 2 and 6.
Predict first
What can be said about its sign on the interval between them?
Correct: It keeps the same sign throughout.
Why: A polynomial can only change sign by passing through zero, and there are no roots in that interval, so its sign is constant there. Which sign it is requires testing one point. This is the observation that makes sign charts work, and §3.7 uses it heavily for rational inequalities.
Faded example
For the polynomial that is the product of x plus 3, x minus 1 squared, and x minus 4.
Fill in the blanks
f(0) = (3)(1)(-4) = -12
Why: Substituting zero into each factor gives 3, then negative 1 squared which is 1, then negative 4. Their product is negative 12. Evaluating the factored form directly is far faster than expanding first, and the y-intercept fixes the vertical scale of the sketch.
Explain it to yourself
The recommended order draws the end behaviour before anything else.
Discussion prompt
Explain why sketching from the ends inward is more reliable than starting in the middle.
Hint: How many ways can a curve be drawn once both ends are fixed?
Answer:
Fixing both ends anchors the curve at the two places where its behaviour is completely determined. There is no choice about the ends, so drawing them first commits to nothing that might be wrong.
The roots then act as constraints on the path between the anchors, and each one's multiplicity says whether the path crosses or turns there. Typically only one shape satisfies all of it, so the sketch draws itself.
Starting in the middle means guessing the curve's direction and discovering at the edges that the ends come out wrong, which requires redrawing. The end behaviour is the one thing you know for certain, so it is the natural place to start.
Section
Section 4
Concept
If a polynomial is negative at one input and positive at another, it must be zero somewhere between them, because its graph has no breaks.
This is the Intermediate Value Theorem, used informally. Its formal statement and the continuity it depends on are §12.3's business, but the idea is available now and it is how a calculator's root-finder works: bracket a sign change, then bisect.
Figure (svg): A curve passing from below the horizontal axis to above it between two marked inputs, with the guaranteed crossing point highlighted between them
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 354-358
Picture it
Negative at one end, positive at the other, and continuous in between.
Figure (svg): A curve passing from below the horizontal axis to above it between two marked inputs, with the guaranteed crossing point highlighted between them
The theorem promises a root somewhere in the shaded strip. It does not say where, and it does not say there is only one — narrowing the interval is what locates it.
Worked example
Two evaluations and a comparison of signs.
\[ \text{Show } f(x)=x^3-x-3 \text{ has a root between } 1 \text{ and } 2. \]
Evaluate at the lower end
Why: One minus one minus three.
\[ f(1) = -3 \]
Evaluate at the upper end
Why: Eight minus two minus three.
\[ f(2) = 3 \]
Compare the signs
Why: One negative, one positive.
Apply the argument
Why: A polynomial has no breaks.
Figure (svg): A curve passing from below the horizontal axis to above it between two marked inputs, with the guaranteed crossing point highlighted between them
\[ f(1)=-3<0 \text{ and } f(2)=3>0, \text{ so a root lies in } (1,2) \]
Verify: narrow it once
Why: At 1.5 the output is 3.375 minus 1.5 minus 3, which is negative 1.125 — still negative. So the root lies between 1.5 and 2, and the interval has halved. Repeating this bisection is exactly how a calculator finds roots numerically, and it converges as far as anyone needs.
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 355-357
Sorting
Sign changes guarantee roots; matching signs guarantee nothing.
Sort into buckets
Sort each observation.
Worked example
The guarantee runs one way only.
\[ \text{Both } f(-2) \text{ and } f(2) \text{ are positive for } f(x)=x^2-1. \text{ Are there roots between?} \]
Evaluate at both ends
Why: Four minus one at each.
\[ \text{both give } 3 \]
Note the signs agree
Why: Both positive.
Ask what that guarantees
Why: Nothing, in either direction.
Check directly
Why: The polynomial factors.
\[ \text{roots at } 1\text{ and } -1 \]
Figure (svg): The solution to Worked example when the same sign proves nothing shown as a ladder of expressions, one row per legal move
\[ \text{Roots at } \pm 1: \text{ two crossings, so the signs at the ends match.} \]
Verify: see why two roots hide
Why: The curve dips below the axis between the two inputs and comes back up, crossing twice. Two sign changes cancel out as far as the endpoints are concerned. So matching signs at the ends means an EVEN number of crossings in between, possibly zero — which is informative, but is not the same as none.
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 357-358
Trap
\[ f(0)=5>0 \text{ and } f(4)=9>0 \;\Longrightarrow\; \text{no roots between } 0 \text{ and } 4 \]
Evaluate at both ends and compare
Why: Both outputs are positive, so no sign change occurs.
The interval is declared root-free.
The guarantee runs only one way. A sign change proves a root exists; the absence of one proves nothing.
The curve could dip below the axis and return, crossing twice. Matching endpoint signs mean an even number of crossings, which includes two, four, or zero.
To rule roots out you need more than two evaluations — a factorisation, or a sign chart, or an argument about the polynomial's minimum. Two samples of a continuous curve say very little about what happens between them.
Prediction
A root is known to lie between 1 and 2, and the midpoint gives a negative output while the lower end was also negative.
Predict first
Where is the root now known to be?
Correct: Between 1.5 and 2.
Why: The sign change must lie between the midpoint and whichever end has the opposite sign, which here is the upper one. The interval has halved, and repeating this narrows it as far as required. This is how root-finding algorithms work, and each step is exactly one evaluation.
Faded example
For the polynomial x cubed minus 2x minus 5, evaluate at 2 and 3.
Fill in the blanks
f(2) = 8 - 4 - 5 = -1, \qquad f(3) = 27 - 6 - 5 = 16
Why: The outputs are negative 1 and 16, which have opposite signs, so a root lies between 2 and 3 — and much closer to 2, since the output there is nearly zero. This is a famous example: the root is about 2.0946, and it was one of the first equations Newton's method was demonstrated on.
Edge cases
The argument requires the graph to have no breaks.
Discussion prompt
Give a function with a break for which the sign-change argument fails, and say why polynomials never have this problem.
Hint: Try the reciprocal rule across zero.
Answer:
The reciprocal rule. At negative 1 it gives negative 1 and at positive 1 it gives 1 — opposite signs — and yet it has no root at all between them. It jumps from far below the axis to far above without ever passing through zero.
The reason is the break at zero, where the function is undefined. The sign change happened across the gap rather than through a crossing.
Polynomials never have this problem because they are defined and continuous everywhere — a consequence of the non-negative whole exponents from §3.3. So the argument always applies to them, and §12.3 will state the general condition under which it applies to anything else.
Section
Section 5
Concept
A graph's roots and their behaviours determine the factors and their exponents. One additional point is needed to determine the leading coefficient.
The last step is often skipped and is where the answer goes wrong. Every vertical stretch of a graph has exactly the same roots and multiplicities, so the factored form without a coefficient describes infinitely many polynomials. The extra point picks out the one you were shown.
Figure (svg): A polynomial graph annotated with the reasoning that recovers its formula: the roots read off as factors, the bouncing root squared, and the leading coefficient found from one extra point
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 358-361
Picture it
Roots to factors, behaviours to exponents, one point to the coefficient.
Figure (svg): A polynomial graph annotated with the reasoning that recovers its formula: the roots read off as factors, the bouncing root squared, and the leading coefficient found from one extra point
The third step is the one with no visual prompt. A sketch with the right roots and the wrong vertical scale looks entirely plausible.
Worked example
Factors first, then the coefficient.
\[ \text{A degree } 3 \text{ graph crosses at } -1 \text{ and bounces at } 2, \text{ with } y\text{-intercept } 8. \]
Write the factors from the roots
Why: One per root.
\[ (x + 1)\text{ and } (x - 2) \]
Assign multiplicities
Why: Crossing is odd, bouncing is even.
\[ (x + 1) (x - 2) ^{2} \]
Check the degree
Why: One plus two is three.
\[ ^\circ 3,\text{ correct} \]
Use the intercept for the coefficient
Why: Substitute zero and solve.
\[ a(1) (4) = 8, a = 2 \]
Figure (svg): A polynomial graph annotated with the reasoning that recovers its formula: the roots read off as factors, the bouncing root squared, and the leading coefficient found from one extra point
\[ f(x)=2(x+1)(x-2)^2 \]
Verify: check the intercept and the degree
Why: At zero: 2 times 1 times 4, which is 8 — the given intercept. The multiplicities sum to 3, matching the stated degree. Without the coefficient 2 the formula would have given an intercept of 4, so the last step genuinely changed the answer.
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 359-360
Faded example
A polynomial has factors x minus 1 and x plus 2, and passes through the point at (0, 6).
Fill in the blanks
a(-1)(2) = 6 \;\Longrightarrow\; -2a = 6 \;\Longrightarrow\; a = -3
Why: Substituting zero into the factors gives negative 1 and 2, whose product is negative 2. Setting a times negative 2 equal to 6 gives a equal to negative 3. The negative coefficient also tells you the end behaviour, which is a free consistency check against the graph.
Worked example
The end behaviour reveals the sign before any arithmetic.
\[ \text{A degree } 4 \text{ graph has both ends down, roots at } 0 \text{ and } 3 \text{ (both bounces).} \]
Read the sign from the ends
Why: Even degree with both ends down.
Write the factors with multiplicities
Why: Two bounces, so two squared factors.
\[ x ^{2}(x - 3) ^{2} \]
Check the degree
Why: Two plus two is four.
\[ ^\circ 4,\text{ correct} \]
Note what is still unknown
Why: The size of a needs another point.
Figure (svg): The solution to Worked example a negative leading coefficient shown as a ladder of expressions, one row per legal move
\[ f(x)=a\,x^2(x-3)^2, \quad a<0 \]
Verify: check the sign is consistent
Why: With a negative and both factors squared, the output is negative or zero everywhere — so the graph sits entirely on or below the axis, touching at 0 and 3. That matches both ends going down and both roots bouncing, so the sign is right. The y-intercept here is zero and gives no information, which is why a different point is needed.
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 360-361
Trap
\[ \text{roots at } -1 \text{ and } 2 \text{ (bounce)}, \; y\text{-intercept } 8 \;\Longrightarrow\; f(x)=(x+1)(x-2)^2 \]
Write the factors from the roots and multiplicities
Why: Both roots are handled correctly and the degree is right.
The formula is reported without any leading coefficient.
That formula gives a y-intercept of 4, not 8. Substituting zero gives 1 times 4, which is 4 — half what the graph shows.
A leading coefficient of 2 is required, and it is found by exactly this substitution: set the formula's value at zero equal to the observed 8 and solve.
Every vertical stretch has the same roots, so the roots alone never determine the polynomial. One more point is always needed, and checking the intercept is the fastest way to catch its absence.
Prediction
Two polynomials have exactly the same roots with the same multiplicities.
Predict first
How can they still differ?
Correct: By a constant multiple, which stretches the graph vertically.
Why: Multiplying by a constant leaves every root and multiplicity untouched while changing every non-zero output. A horizontal shift would move the roots, and a different degree would change the multiplicities, so neither is possible with the roots fixed. This is exactly why the extra point is needed.
Matching
Different features supply different parts of the formula.
Match the pairs
Why: The roots and their behaviours build the factored form up to a constant, and the end behaviour confirms the degree's parity and the coefficient's sign. Only the extra point pins down the coefficient's size, which is why it is the step that cannot be skipped.
Real world
Recovering a formula from observed behaviour is what curve fitting does.
Discussion prompt
An engineer knows a quantity is zero at three specific inputs and is measured at one other. Why is that enough to fit a cubic?
Hint: How many unknowns does a cubic have, and how many facts were given?
Answer:
Three roots give three factors, which determines the cubic up to a constant multiple. That is three of the four unknowns in a general cubic settled at once.
The one measured value supplies the fourth constraint, fixing the leading coefficient. Four facts, four unknowns, one answer.
This is why root information is so valuable in fitting: each root is worth a whole factor, which is far more constraining than an ordinary data point. Knowing where something is zero tells you more about its shape than knowing where it passes through some arbitrary value.
Comparison
Fill the blanks from memory. The parity decides the behaviour and the size decides the flatness.
Comparison matrix
| multiplicity 1 | multiplicity 2 | multiplicity 3 | |
|---|---|---|---|
| parity | odd | even | odd |
| at the axis | crosses | bounces | crosses |
| sign of the output | changes | does not change | changes |
| shape of approach | straight, at an angle | like a parabola's vertex | flattened, like a cubic |
| is it a turning point | no | yes | no |
The third row is the reason for the second. The graph crosses exactly when the output's sign changes, and the parity of the multiplicity is what decides that.
Pattern
Six steps, and after them the sketch is determined.
Step 2's sum check is worth doing every time. A set of multiplicities that does not add to the degree means a factor has been missed or an exponent misread, and it is far cheaper to catch that here than after the sketch is drawn.
OpenStax Algebra and Trigonometry 2e, §5.3 Graphs of Polynomial Functions §5.3
Check
Odd crosses, even bounces.
Check your understanding
At the root x = 3 of the polynomial (x - 3)^4 (x + 1), what does the graph do?
Answer: A
Why: The multiplicity is 4, which is even, so the output does not change sign as the input passes through 3. The curve touches the axis and returns, and the flattening from the high multiplicity makes the touch very shallow.
Check
Add the multiplicities.
Check your understanding
What is the degree of the polynomial (x - 1)^2 (x + 5)^3 (x - 2)?
Answer: A
Why: The exponents 2, 3 and 1 sum to 6, which is the degree. Each factor contributes its own exponent, so the degree is the total rather than the count of brackets or the largest exponent.
Check
The guarantee runs one way.
Check your understanding
A polynomial gives outputs 4 and 9 at two inputs. What follows about roots between them?
Answer: A
Why: Both outputs are positive, so there is no sign change and the argument gives no conclusion. The curve may stay above the axis, or it may dip below and return, crossing an even number of times. More information is needed to decide.
Real world
Bracketing a sign change is how numerical root-finders work, and they are everywhere.
Discussion prompt
A calculator finds a root of an equation it cannot solve algebraically. What is it most likely doing?
Hint: What can it compute easily, and what does a sign change guarantee?
Answer:
It can evaluate the function at any input cheaply, even when it cannot solve the equation. So it searches for two inputs where the outputs have opposite signs, which brackets a root.
Then it bisects: evaluate at the midpoint, keep whichever half still shows a sign change, and repeat. Each step halves the interval, so about fifty steps pins the root down to the limits of the machine's precision.
More sophisticated methods converge faster, but they all rest on the same guarantee this section states informally: a continuous function that changes sign must pass through zero. It is one of those results that is obvious enough to be easy to overlook and load-bearing enough to be worth naming.
Commit first
State your confidence along with your answer.
Predict first
A degree 5 polynomial has roots at 1 and 2 only, and bounces at 1. What is the multiplicity at 2?
Correct: 3, since the multiplicities must sum to 5.
Why: The bounce at 1 means an even multiplicity, and the smallest is 2. With only two roots and a total of 5, the remaining multiplicity must be 3 — which is odd, so the graph crosses at 2 with a flattening. The sum check does the whole job here, and it is why that check is worth performing every time.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why a squared factor makes the graph bounce rather than cross, without simply stating the rule.
Hint: What is the sign of a squared quantity, on either side of zero?
Answer:
Near the root, only that factor changes sign — every other factor is nowhere near its own root and keeps whatever sign it has. So the product's sign is decided by that one factor.
A squared factor is never negative, whichever side of the root you are on. So the product keeps the same sign on both sides, the output never crosses zero, and the graph touches and turns back.
The contrast makes it clear: an unsquared factor is negative on one side and positive on the other, so the product genuinely flips sign and the curve passes through. The parity of the exponent is exactly the question of whether the sign change survives.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second is the section's central idea and everything else depends on it. The fourth is where the most marks are quietly lost, because the leading coefficient step has no visual prompt and its omission produces an answer that looks entirely correct.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write down a factored polynomial with at least three distinct roots, one of them squared and one cubed. Sketch it: draw the end behaviour first, mark each root with its behaviour, plot the y-intercept, and join up. Then, beside the sketch, write the three facts you would need to recover this formula if you had only been shown the picture.
If your third fact is a point off the axis, you have the step that the roots alone can never supply.
Recap
Five things, and the second one is what turns a list of roots into a picture.
| if you remember one thing | it should be this |
|---|---|
| about multiplicity | odd crosses, even bounces, and higher flattens either way |
| about factoring | take out the common factor and record the root at zero |
| about sign changes | they guarantee a root; matching signs guarantee nothing |
| about recovering a formula | the roots never determine the leading coefficient |
Section 3.5 supplies the tool this one assumed: a way to factor a polynomial whose roots are not obvious, by dividing out a factor you already know.
OpenStax, Precalculus, §3.4 Graphs of Polynomial Functions §3.4, pp. 335-361 — everything on these slides traces back here
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