HSA.APR.B.3: Finding Zeros of Polynomials and Sketching Rough Graphs
In plain English: HSA.APR.B.3 is the Common Core algebra standard that asks students to find the zeros of a polynomial from a suitable factorization and use them to sketch a rough graph of the function. Each factor x - r gives a zero at r, and the zeros, whether the graph crosses or touches at each one, the y-intercept and the end behavior give the shape. It is usually taught in Algebra II.
Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.
Common Core State Standards for Mathematics · Domain: Arithmetic with Polynomials and Rational Expressions (APR) · Cluster: Understand the relationship between zeros and factors of polynomials Also written as HSA-APR.B.3 or A-APR.3 · Official standard
Students start from a polynomial in factored form, or one they can factor with familiar tools (a common factor, grouping, a difference of squares, a quadratic trinomial, or a factor they are given), and use the zero product property to list its zeros. They then build a rough graph from four pieces of information: the zeros, whether the graph crosses or touches at each zero, the y-intercept, and the end behavior read from the leading term.
A "rough graph" means the right shape in the right places, not exact turning points. Students confirm the shape with a sign chart, testing one value between each pair of zeros, and check their sketches with technology only after drawing them by hand.
Learning Objectives
By the end of this lesson, students will be able to:
Identify the zeros of a polynomial written as a product of linear factors, including factors such as 2x - 1 and x
Factor a polynomial using a common factor, grouping, a difference of squares or a known factor in order to find its zeros
Use the sign of the polynomial between consecutive zeros to decide where the graph is above or below the x-axis
Tell from the exponent on a repeated factor whether the graph crosses the x-axis or touches it and turns
Sketch a rough graph that shows the zeros, y-intercept and end behavior
Prior Knowledge Required
Students should already be comfortable with:
Factoring quadratics and pulling out common factors HSA.SSE.A.2
Solving quadratic equations by factoring and the zero product property HSA.REI.B.4
The Remainder Theorem: p(a) = 0 exactly when x - a is a factor HSA.APR.B.2
Reading x-intercepts and y-intercepts from a graph HSF.IF.B.4
Students work alone for 4 minutes, then compare with a partner.
Warm-Up Prompt
"Let f(x) = (x + 2)(x - 1)(x - 3). (1) Find f(-3), f(0), f(2) and f(4). (2) For which three values of x is f(x) = 0? (3) Plot the seven points you now have and connect them with a smooth curve."
Answers: f(-3) = -24, f(0) = 6, f(2) = -4, f(4) = 18, and f(x) = 0 at x = -2, 1 and 3. Ask: "How could you have found the three zeros without plugging anything in?" and "Why does the graph switch from below the axis to above it at each zero?" Keep the sketch posted; it becomes Diagram 1.
Direct Instruction20 minutes
Present a four-step routine for a rough graph, then work five examples, each showing a different factoring situation:
Factor completely and list the zeros. A product is 0 exactly when one factor is 0, so each factor x - r gives the zero r. The factor x gives the zero 0, and 2x - 1 gives the zero 1/2.
Check each zero's exponent. An odd exponent (like (x - 2)¹) means the graph crosses the x-axis there. An even exponent (like (x + 1)²) means the graph touches the axis and turns back, because the squared factor never changes sign.
Find the y-intercept and the end behavior. The y-intercept is p(0). Multiply the leading terms of the factors to get the leading term: its degree and sign tell how the graph behaves at the far left and far right.
Sketch and confirm with a sign chart. Test one x-value in each interval between zeros. The sign tells you whether that piece of the graph is above or below the x-axis.
Already factored cubic
f(x) = (x + 2)(x - 1)(x - 3)
Equation: Zeros -2, 1, 3 (all cross); y-intercept 6; leading term x³, so falls to the left and rises to the right
p(x) = -(x - 1)²(x + 3): zeros -3 (crosses) and 1 (touches); y-intercept -3; rises left, falls right
Look for boards that show the right zeros but the wrong end behavior. That usually means the student ignored the negative sign in front, or thought every graph starts low on the left.
Independent Practice10 minutes
Students complete three problems alone: one that must be factored by grouping, one with a repeated factor, and one quartic. For each, they write the zeros with their multiplicities, the y-intercept, the end behavior and a sign chart before sketching.
Closure5-10 minutes
Exit ticket: "Sketch a rough graph of p(x) = x²(x - 3). Label the zeros and say what the graph does at each one." Answer: the graph touches at 0 (x² never changes sign) and crosses at 3; the leading term is x³, so it falls to the left and rises to the right; it is at or below the axis for x < 3 and above the axis for x > 3.
Differentiation Strategies
For Struggling Students
Give a graphic organizer with four boxes: zeros, crosses or touches, y-intercept, end behavior
Start with polynomials already in factored form before adding factoring
Have students write "x + 3 = 0, so x = -3" for every factor until the sign flip is automatic
For Advanced Students
Sketch p(x) = (x - 1)³(x + 2) and describe how the flattening at x = 1 differs from a simple crossing
Write a polynomial in factored form whose graph crosses at -3, touches at 2 and falls on both ends, for example -(x + 3)(x - 2)²(x + 1) or another valid answer; explain each choice
Explain why a polynomial of degree 3 must have at least one real zero, using end behavior
Assessment Guidance
What to Look For
Check three things in every sketch: the zeros are in the right places with the right sign (x + 3 gives -3), the graph touches rather than crosses at zeros from even-exponent factors, and the ends match the sign and degree of the leading term. A sketch with correct zeros but wrong ends usually means the student never multiplied the leading terms.
02
Classroom Activities
3 Activities
1
Graph Match-Up
20 minGroups of 3-4
Groups get six polynomial cards in factored form and six unlabeled rough-graph cards. They match each polynomial to its graph and must name the feature that rules out the closest wrong graph. The set includes near-miss pairs that differ only in end behavior or in crossing versus touching.
Card Set
(x + 2)(x - 1)(x - 3) and -(x + 2)(x - 1)(x - 3): same zeros, opposite end behavior
(x + 2)²(x - 1) and (x + 2)(x - 1)²: touch and cross swapped
x(x - 2)(x + 2) and x²(x - 2)(x + 2): degree 3 versus degree 4, and a touch at 0
Procedure
Groups have 10 minutes to match and write one justification per match, such as "touches at -2 because of the squared factor"
Each group checks one match of another group's using the y-intercept as a quick test
Debrief: which feature helped the most? Which pairs were hardest to tell apart?
Modification for Distance Learning
Put the cards on a shared slide and have groups drag each equation next to its graph. Reveal the answers with a graphing app one pair at a time.
2
Factor, Then Sketch Relay
20 minGroups of 4
Each group gets a polynomial that is not yet factored. Student 1 factors it, Student 2 lists the zeros with multiplicities, Student 3 finds the y-intercept and end behavior, and Student 4 draws the sketch. Each student checks the previous step before starting their own, and the roles rotate every round.
Round Polynomials
Round 1: x³ - 9x = x(x - 3)(x + 3) (common factor, then difference of squares)
Round 4: 2x³ + x² - 8x - 4 = (2x + 1)(x - 2)(x + 2) (grouping, with a zero at -1/2)
Discussion Questions
Which factoring method did each round need, and what clue in the polynomial pointed to it?
In Round 4, why is the zero -1/2 and not -1?
How can you check the whole relay in 10 seconds? (Compare the y-intercept on the sketch with p(0).)
3
Predict, Then Verify
15 minIndividual then share
Students sketch four rough graphs by hand, then graph each polynomial with technology and write down anything they got wrong. The written list of errors is what they hand in, and it shows the teacher which feature each student misses most.
Polynomials to Sketch
(x - 3)(x + 1)(x + 4)
-x(x - 2)²
x³ - 2x² + 4x - 8 = (x - 2)(x² + 4): only one x-intercept
-(x² - 9)(x² - 1)
Reflection Prompts
For each graph, what did your sketch get right: zeros, crossing or touching, y-intercept, ends?
Why can't a rough sketch show the exact height of the turning points?
Why does the factor x² + 4 add no x-intercepts?
Extension Variation: Write the Equation
Give students a rough graph with labeled zeros and a y-intercept, and ask for a factored polynomial that fits. This reverses the standard's direction and works well as an extension for students who finish early.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: From Zeros to a Rough Graph
The zeros -2, 1 and 3 split the x-axis into four intervals. One test value per interval gives the sign of f(x), and the sign says whether the graph is above or below the axis there. The y-intercept f(0) = 6 and the leading term x³ complete the sketch. Drawn to scale.
Diagram 2: Touching vs. Crossing, and End Behavior
Left: the squared factor (x + 1)² never changes sign, so the graph touches the axis at -1 and turns back, while the single factor (x - 2) makes it cross at 2. Right: a degree-4 polynomial with a negative leading coefficient falls on both ends. Both graphs are drawn to scale.
04
Homework Assignment
~30 min
HSA.APR.B.3 Homework: Zeros and Rough Graphs
Directions: For each polynomial, (a) factor it completely if it is not already factored, (b) list the zeros and say whether the graph crosses or touches at each one, (c) find the y-intercept, (d) describe the end behavior, and (e) sketch a rough graph with the zeros and y-intercept labeled.
Part 1: Polynomials in Factored Form (Problems 1-3)
f(x) = (x - 4)(x + 1)(x + 5)
g(x) = -3x(x - 4)(x + 1)
h(x) = (x - 2)²(x + 3)
Part 2: Factor First (Problems 4-6)
p(x) = x³ - 25x
p(x) = x³ - 3x² - 4x + 12 (hint: group the first two terms and the last two terms)
p(x) = x³ + 4x² + x - 6, given that x + 2 is a factor. Use division to find the other factor.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Factoring
Completely factored, all work shown
Partly factored or one error
Not factored
Zeros and Multiplicity
All zeros correct, crossing or touching correct
One zero or one crossing/touching decision wrong
Several zeros wrong or missing
Intercept and End Behavior
y-intercept and both ends correct
One of the two wrong
Both wrong or missing
Sketch
Sketch matches all features and is labeled
Sketch has one feature wrong or missing labels
No sketch, or sketch does not match the work
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
What are the zeros of f(x) = (x - 3)(x + 5)?
Answer: B
Set each factor to 0: x - 3 = 0 gives 3, and x + 5 = 0 gives -5. Choice A reverses both signs, a common error. Choice D is f(0), the y-intercept, not a zero.
Question 2 of 20 · Multiple Choice
What are the zeros of p(x) = x(x + 4)(2x - 1)?
Answer: C
x = 0 gives 0, x + 4 = 0 gives -4, and 2x - 1 = 0 gives 1/2. Choice B forgets that the factor x gives the zero 0. Choice D solves 2x - 1 = 0 as x = 1 by ignoring the coefficient 2.
Question 3 of 20 · Multiple Choice
Factor p(x) = x³ - 16x to find its zeros. What are they?
Answer: A
x³ - 16x = x(x² - 16) = x(x - 4)(x + 4), so the zeros are 0, 4 and -4. Choice B drops the common factor x after dividing it out. Choice D treats x² - 16 = 0 as x = ±16 instead of taking the square root.
Question 4 of 20 · Multiple Choice
Which describes the end behavior of f(x) = -(x - 1)(x + 2)(x - 4)?
Answer: D
Multiply the leading terms: -(x)(x)(x) = -x³. A cubic with a negative leading coefficient rises to the left and falls to the right. Choice A is the behavior of +x³, the result of ignoring the negative sign in front. Choices B and C describe even-degree polynomials.
Question 5 of 20 · Multiple Choice
What does the graph of p(x) = (x - 3)²(x + 1) do at x = 3?
Answer: B
The factor (x - 3)² is never negative, and x + 1 is positive near 3, so p(x) has the same sign on both sides of 3: the graph touches the axis and turns. Choice A would be true at x = -1, where the factor has exponent 1. Polynomial graphs have no gaps, so choice C is impossible.
Question 6 of 20 · Multiple Choice
What is the y-intercept of f(x) = (x - 2)(x + 3)(x - 1)?
Answer: C
f(0) = (-2)(3)(-1) = 6. Choice A loses one of the two negative signs. Choice B confuses the y-intercept with a zero.
Question 7 of 20 · Multiple Choice
Which polynomial has zeros -1, 2 and 4 and no other zeros?
Answer: B
A zero r comes from the factor x - r, so -1, 2 and 4 give (x + 1), (x - 2) and (x - 4). Choice A has every sign reversed and gives zeros 1, -2 and -4. Choice D adds the factor x, which makes 0 a fourth zero.
Question 8 of 20 · Multiple Choice
For f(x) = (x + 4)(x + 1)(x - 2), what is the sign of f(x) when -1 < x < 2?
Answer: D
Test x = 0: (4)(1)(-2) = -8, so the graph is below the x-axis on that whole interval. The sign can change only at a zero, and 0 is not a zero of f, so choice B is wrong. Choice C comes from checking only the first two factors.
Question 9 of 20 · Multiple Choice
What are the zeros of p(x) = x³ - 3x² - x + 3?
Answer: A
Group: x²(x - 3) - 1(x - 3) = (x - 3)(x² - 1) = (x - 3)(x - 1)(x + 1). The zeros are 3, 1 and -1. Choice C stops after the first factor and forgets that x² - 1 also factors. Choice B has the sign of the zero from x - 3 reversed.
Question 10 of 20 · Multiple Choice
How many x-intercepts does the graph of p(x) = x⁴ - 13x² + 36 have?
Answer: C
x⁴ - 13x² + 36 = (x² - 4)(x² - 9) = (x - 2)(x + 2)(x - 3)(x + 3), so there are 4 x-intercepts: ±2 and ±3. Choice A counts the two solutions x² = 4 and x² = 9 but forgets that each gives a positive and a negative x.
Question 11 of 20 · Multiple Choice
A graph crosses the x-axis at -3, touches the x-axis at 2, falls to the left and rises to the right. Which function could it be?
Answer: D
Crossing at -3 needs (x + 3) to an odd power; touching at 2 needs (x - 2) squared; falls left and rises right needs a positive leading term x³. Only choice D has all three. Choice A swaps touching and crossing. Choice B has the right zeros but the negative sign flips the end behavior. Choice C has the zeros at the wrong signs.
Question 12 of 20 · Multiple Choice
Which statement about g(x) = x²(x - 5) is true?
Answer: B
x² gives the zero 0 with multiplicity 2, so the graph touches there; x - 5 gives the zero 5 with multiplicity 1, so it crosses there. Choice C forgets that x² = 0 gives a zero. Choice D attaches the multiplicities to the wrong zeros.
Question 13 of 20 · Multiple Choice
Given that x - 2 is a factor of p(x) = x³ - 2x² - 9x + 18, what are all the zeros of p?
Answer: A
Divide (or group): x³ - 2x² - 9x + 18 = x²(x - 2) - 9(x - 2) = (x - 2)(x² - 9) = (x - 2)(x - 3)(x + 3). Zeros: 2, 3 and -3. Choice B takes the 9 from x² - 9 as a zero instead of solving x² = 9. Choice C gets the sign of the zero from x - 2 wrong.
Question 14 of 20 · Multiple Choice
How many real zeros does p(x) = x³ + x have?
Answer: C
x³ + x = x(x² + 1). The factor x gives the zero 0, but x² + 1 is at least 1 for every real x, so it adds no real zeros. The graph crosses the x-axis only once, at the origin. Choice A assumes a cubic always has three real zeros.
Question 15 of 20 · Short Answer
Sketch a rough graph of f(x) = (x + 3)(x - 1)(x - 4). List the zeros, the y-intercept, the end behavior and the sign of f(x) on each interval.
Zeros: -3, 1, 4 (each crosses). y-intercept: f(0) = (3)(-1)(-4) = 12. End behavior: leading term x³, so falls to the left and rises to the right. Signs: negative for x < -3, positive between -3 and 1, negative between 1 and 4, positive for x > 4 (test values f(-4) = -40, f(0) = 12, f(2) = -10, f(5) = 32).
Question 16 of 20 · Short Answer
Factor p(x) = x³ + x² - 6x completely, list its zeros, and sketch a rough graph.
p(x) = x(x² + x - 6) = x(x + 3)(x - 2). Zeros -3, 0 and 2, each crossing. The y-intercept is 0 because 0 is a zero. The leading term x³ means the graph falls to the left and rises to the right: below the axis for x < -3, above between -3 and 0, below between 0 and 2, above for x > 2.
Question 17 of 20 · Short Answer
An open box is made from a 12-inch by 8-inch sheet of cardboard by cutting squares of side x inches from each corner and folding up the sides. Its volume is V(x) = x(12 - 2x)(8 - 2x). Find the zeros of V, sketch a rough graph, and say which part of the graph makes sense for the box.
Setting each factor to 0 gives x = 0, x = 6 and x = 4. The leading term is x · (-2x) · (-2x) = 4x³, so the graph falls to the left and rises to the right, with V positive on 0 < x < 4, negative on 4 < x < 6, and positive again for x > 6. Only 0 < x < 4 fits the box: at x = 4 the 8-inch side is used up, and beyond that a side length would be negative. The positive values for x > 6 come from two negative side lengths and have no physical meaning. For example, V(2) = 2 · 8 · 4 = 64 cubic inches.
Question 18 of 20 · Short Answer
A student says the graph of f(x) = (x + 2)²(x - 4) crosses the x-axis at x = -2. Explain the error using the sign of f(x) on each side of -2.
Test a value on each side: f(-3) = (1)(-7) = -7 and f(-1) = (1)(-5) = -5. Both are negative, so the graph is below the axis on both sides of -2: it touches the x-axis at -2 and turns back. The squared factor (x + 2)² is never negative, so it cannot make f change sign. The graph does cross at 4, where the factor x - 4 has exponent 1.
Question 19 of 20 · Short Answer
Sketch a rough graph of h(x) = -x⁴ + 10x² - 9. Factor first.
h(x) = -(x⁴ - 10x² + 9) = -(x² - 1)(x² - 9) = -(x - 1)(x + 1)(x - 3)(x + 3). Zeros -3, -1, 1 and 3, each crossing. y-intercept h(0) = -9. The leading term is -x⁴, so both ends fall. Signs: negative for x < -3, positive on -3 < x < -1, negative on -1 < x < 1, positive on 1 < x < 3, negative for x > 3 (for example h(2) = 15).
Question 20 of 20 · Short Answer
Factor p(x) = x³ + 3x² + 9x + 27 by grouping, and explain why its graph has only one x-intercept.
p(x) = x²(x + 3) + 9(x + 3) = (x + 3)(x² + 9). The factor x + 3 gives the zero -3. The factor x² + 9 is at least 9 for every real x, so it is never 0 and adds no x-intercepts. The graph crosses the x-axis only at x = -3, has y-intercept 27, and falls to the left and rises to the right.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What counts as a "rough graph" for this standard?
A rough graph shows the right shape in the right places: the zeros on the x-axis, whether the graph crosses or touches at each zero, the y-intercept, and the end behavior. It does not need exact turning points. If the sketch is above and below the x-axis on the right intervals, it meets the standard.
What does "when suitable factorizations are available" mean?
The polynomial is either already factored or can be factored with tools students know: a common factor, grouping, a difference of squares, a quadratic trinomial, a quadratic in x² such as x⁴ - 5x² + 4, or a factor they are given. Students are not expected to find zeros of polynomials that do not factor nicely; that is a job for technology.
Why does a squared factor make the graph bounce off the x-axis?
A factor like (x - 2)² is never negative, so it does not change sign as x passes 2. The other factors keep their signs near 2 too, so p(x) has the same sign on both sides and the graph touches the axis and turns back. With an odd exponent the factor does change sign, so the graph crosses.
How do I find the end behavior quickly?
Multiply the leading terms of the factors. For -2x(x - 3)(x + 2) that is -2 · x · x · x = -2x³. Then read the ends from it: odd degree means the ends go in opposite directions, even degree means they go the same way, and a negative leading coefficient flips the picture you would get for a positive one.
Why can't we find the exact turning points?
The height and location of a turning point are not determined by the zeros alone, and finding them exactly takes calculus. For this standard students only need the turning points in the right intervals. A graphing app can give their approximate coordinates when needed.
What happens with a factor like x² + 4 that does not factor?
x² + 4 is always positive for real x, so it has no real zeros and adds no x-intercepts. It still affects the degree, the end behavior and the y-intercept. For example, (x - 2)(x² + 4) is a cubic with only one x-intercept.
What mistakes do students often make on HSA.APR.B.3?
Reading the zero of x + 3 as 3 instead of -3
Forgetting the zero 0 when a factor of x is pulled out
Reading the zero of 2x - 1 as 1 instead of 1/2
Ignoring a negative leading coefficient, which flips the end behavior
Drawing a crossing at a zero that comes from a squared factor
Stopping factoring too early, as in (x + 5)(x² - 49)
How does this standard connect to the Remainder Theorem?
HSA.APR.B.2 says p(a) = 0 exactly when x - a is a factor. That is why each factor gives a zero, and why a known zero can be used to divide and find the remaining factors, as in homework Problem 6.
Is HSA.APR.B.3 on the SAT?
Yes, in the Advanced Math domain. Questions often give a polynomial in factored form and ask for its x-intercepts, or show a graph and ask which factored expression could define it. Knowing that the factor x - r corresponds to the zero r is the key skill.
Should students use a graphing calculator for this?
Have students sketch by hand first, then use technology to check. The standard is about reasoning from the factors to the graph, and a calculator skips that reasoning. Comparing a hand sketch with the calculator graph is a quick way for students to find their own errors.
07
Related Standards
6 standards
These standards connect to HSA.APR.B.3: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSA.SSE.A.2Prerequisite
Use the structure of an expression to identify ways to rewrite it
