Lesson Map
The attached Chapter 3 notes establish five lessons: multiplying polynomials; factoring trinomials with a = 1; factoring trinomials with a > 1; factoring special cases; and polynomial division. The review then mixes those skills instead of treating them as isolated procedures.
| Lesson | Section | Focus | You should be able to… |
|---|---|---|---|
| Lesson 1 | 3.1 | Multiplying polynomials | Use the distributive property to multiply a monomial by a polynomial, two binomials, and a binomial by a larger polynomial; combine like terms correctly. |
| Lesson 2 | 3.2 | Factoring trinomials, a = 1 | Factor a GCF first when needed, then find two integers whose product is c and whose sum is b. |
| Lesson 3 | 3.3 | Factoring trinomials, a > 1 | Use the AC method: multiply a·c, split the middle term, and factor by grouping. |
| Lesson 4 | 3.4 | Special factoring cases | Recognize and factor differences of squares and sums/differences of cubes, including expressions that require a GCF first. |
| Lesson 5 | 3.5 | Polynomial division | Divide with long division and use synthetic division when the divisor is linear in the required form. |
| Performance Task | Unit synthesis | Build Polynomial Farm | Stake out a polynomial field layout with measuring tapes and flags, then check the predicted perimeter, area, factored dimensions, and doubling against real measurements. |
How This Unit Is Built
Every support below is a direct link. Use this as the unit home base instead of scrolling to hunt for a resource.
Content Summary
The goal is not to memorize five unrelated procedures. Multiplication expands factors into a polynomial. Factoring reverses that process. Division asks how much of one polynomial fits into another. The best check is often to reverse the operation.
Multiplying Polynomials
The distributive property is the engine behind every multiplication problem in this lesson. Each term in one factor must multiply every term in the other factor. Exponents change only when powers with the same base are multiplied; coefficients multiply normally.
Monomial × polynomial
4x²(3x³ − 2x + 5) = 12x⁵ − 8x³ + 20x²
Distribute 4x² to all three terms. The first term is 4·3·x²·x³ = 12x⁵. Notice that x² and x³ multiply to x⁵; the exponents add.
Binomial × binomial
(2x + 3)(x − 4) = 2x² − 8x + 3x − 12 = 2x² − 5x − 12
FOIL is a useful memory device for two binomials, but distribution is the actual rule. That matters because FOIL does not generalize to a binomial times a trinomial.
Binomial × larger polynomial
(x + 2)(3x² − x + 4)
= x(3x² − x + 4) + 2(3x² − x + 4)
= 3x³ + 5x² + 2x + 8
Factoring Trinomials When a = 1
A quadratic trinomial in standard form is ax² + bx + c. When a = 1, factoring asks for two numbers that multiply to c and add to b.
x² + 7x + 12 = (x + 3)(x + 4)
Why? 3·4 = 12 and 3 + 4 = 7. The signs can be predicted before searching: if c is positive, the two signs match; if c is negative, the signs differ. Their sum must still match b.
Always check for a GCF first
3x² + 18x + 24 = 3(x² + 6x + 8) = 3(x + 2)(x + 4)
Factoring is not complete until every factor is factored as far as the course expects. Missing the GCF at the start is one of the easiest ways to leave an answer incomplete.
Factoring Trinomials When a > 1: The AC Method
Now the first coefficient matters. For ax² + bx + c, find two integers that multiply to a·c and add to b. Use them to split the middle term, then factor by grouping.
6x² + 7x + 2
- Compute a·c = 6·2 = 12.
- Find two numbers that multiply to 12 and add to 7: 3 and 4.
- Split the middle term: 6x² + 3x + 4x + 2.
- Group: 3x(2x + 1) + 2(2x + 1).
- Factor the repeated binomial: (3x + 2)(2x + 1).
If all terms share a GCF, remove it before doing AC. This makes the numbers smaller and prevents an incomplete final answer.
Factoring Special Cases
Special cases are recognition problems. Before using a general method, inspect the number of terms, signs, exponents, and whether the terms are perfect squares or perfect cubes.
Difference of two squares
a² − b² = (a + b)(a − b)
25x² − 49 = (5x + 7)(5x − 7)
This pattern requires a difference. A sum such as a² + b² does not factor with the difference-of-squares rule.
Sum and difference of cubes
a³ + b³ = (a + b)(a² − ab + b²)
a³ − b³ = (a − b)(a² + ab + b²)
A useful sign pattern is: the first binomial keeps the original sign; the middle term of the trinomial uses the opposite sign; the last term is always positive.
x³ − 64 = (x − 4)(x² + 4x + 16)
Division of Polynomials
Polynomial long division follows the same cycle as numerical long division: divide, multiply, subtract, bring down, repeat. Keep the dividend in descending powers and insert zero placeholders for missing powers.
(x² + 5x + 6) ÷ (x + 2) = x + 3
First divide x² by x to get x. Multiply x(x + 2), subtract, and continue. A zero remainder means the divisor is a factor of the dividend.
Synthetic division
Synthetic division is a shortcut for division by a linear divisor of the form x − c (equivalently x + a after rewriting its zero). Use only the coefficients, including a zero coefficient for any missing power.
(2x³ − 3x² − 8x + 12) ÷ (x − 2)
Use c = 2 in the synthetic setup. The resulting quotient is 2x² + x − 6 with remainder 0.
Lesson Videos
Use these after the written explanation when you want to see the procedure carried out in real time. Each card is matched to one lesson rather than being a general playlist.
Multiplying Polynomials
Khan Academy example emphasizing distribution and combining like terms.
Factoring Trinomials: a = 1
Use the linked lesson for a second explanation of factoring monic trinomials by finding the factor pair whose product is c and sum is b.
Factoring Trinomials: a > 1
Factoring by grouping after splitting the middle term — the same structure used by the AC method.
Special Factoring Patterns
Short Khan Academy explanation of the difference-of-squares pattern. The written lesson above also includes the sum and difference of cubes.
Long & Synthetic Division
A worked comparison of the two polynomial-division procedures.
Vocabulary
Polynomial
An expression made of terms with variables raised to nonnegative integer powers.
Term
A number, variable, or product separated from other terms by addition or subtraction.
Coefficient
The numerical factor multiplying a variable term.
Degree
The greatest exponent of the variable in a one-variable polynomial.
Standard form
Terms written in descending order of exponent, such as ax² + bx + c.
Factor
An expression multiplied by another expression to produce a product.
Greatest common factor (GCF)
The largest factor shared by every term. Check for it before other factoring methods.
Trinomial
A polynomial with exactly three terms.
Perfect square
A number or monomial that can be written as another expression squared.
Perfect cube
A number or monomial that can be written as another expression cubed.
Dividend / divisor / quotient
The expression being divided, the expression dividing it, and the result of the division.
Remainder
What is left after polynomial division. A remainder of zero shows the divisor is a factor.
Synthetic division
A coefficient-based shortcut for division by an appropriate linear divisor.
Sample Problems
These are new problems written to match the skills and difficulty of the attached Chapter 3 materials; they are not copied from the review packet.
Distribute a monomial
Simplify: −4x³(3x⁴ − 2x² + 5).
Show solution
−12x⁷ + 8x⁵ − 20x³.
Multiply two binomials
Simplify: (5x − 3)(2x + 7).
Show solution
10x² + 35x − 6x − 21 = 10x² + 29x − 21.
Multiply a binomial and trinomial
Simplify: (2x + 1)(3x² − 4x + 6).
Show solution
6x³ − 8x² + 12x + 3x² − 4x + 6 = 6x³ − 5x² + 8x + 6.
Factor with a = 1
Factor completely: x² + 11x + 24.
Show solution
3·8 = 24 and 3 + 8 = 11, so (x + 3)(x + 8).
Factor with opposite signs
Factor completely: x² − 4x − 21.
Show solution
−7·3 = −21 and −7 + 3 = −4, so (x − 7)(x + 3).
GCF first
Factor completely: 4x² + 28x + 40.
Show solution
4(x² + 7x + 10) = 4(x + 5)(x + 2).
AC method
Factor completely: 8x² + 14x + 3.
Show solution
ac = 24; 12 + 2 = 14. 8x² + 12x + 2x + 3 = 4x(2x + 3) + 1(2x + 3) = (4x + 1)(2x + 3).
AC method with a negative constant
Factor completely: 6x² − x − 12.
Show solution
ac = −72; 8 and −9 add to −1. 6x² + 8x − 9x − 12 = 2x(3x + 4) − 3(3x + 4) = (2x − 3)(3x + 4).
GCF before AC
Factor completely: 10x² + 35x + 25.
Show solution
5(2x² + 7x + 5) = 5(2x + 5)(x + 1).
Difference of squares
Factor completely: 36x² − 121.
Show solution
(6x)² − 11² = (6x + 11)(6x − 11).
Difference of squares with two variables
Factor completely: 49a² − 25b².
Show solution
(7a + 5b)(7a − 5b).
Sum of cubes
Factor completely: x³ + 27.
Show solution
x³ + 3³ = (x + 3)(x² − 3x + 9).
Difference of cubes
Factor completely: 8y³ − 125.
Show solution
(2y)³ − 5³ = (2y − 5)(4y² + 10y + 25).
Polynomial long division
Divide: (x² + 7x + 10) ÷ (x + 5).
Show solution
The quotient is x + 2 with remainder 0.
Long division with a remainder
Divide: (2x² + 5x + 1) ÷ (x + 2).
Show solution
Quotient 2x + 1, remainder −1, so 2x + 1 − 1/(x + 2).
Synthetic division
Use synthetic division: (x³ − 4x² − x + 4) ÷ (x − 4).
Show solution
Use 4 with coefficients 1, −4, −1, 4. The quotient is x² − 1 with remainder 0.
Equations & Rules
Everything the unit needs in one reference section.
a(b + c) = ab + ac. Every term must be distributed.
xmxn = xm+n.
ax² + bx + c.
Before any factoring method, ask whether every term shares a numerical or variable factor.
Find m,n so mn = c and m + n = b; then x² + bx + c = (x + m)(x + n).
Find m,n so mn = ac and m + n = b; split bx into mx + nx; factor by grouping.
a² − b² = (a + b)(a − b).
a³ + b³ = (a + b)(a² − ab + b²).
a³ − b³ = (a − b)(a² + ab + b²).
Divide → multiply → subtract → bring down → repeat.
Use for a linear divisor x − c; use c in the synthetic setup.
Dividend = divisor·quotient + remainder.
Unit 3 Review Guide
The attached review emphasizes multiplication, complete factoring, special cases, and polynomial division. The practice below targets the same skills with different coefficients and expressions.
Skills checklist
- Distribute a monomial across a polynomial.
- Multiply binomials and larger polynomials.
- Combine like terms and write the result in standard form.
- Factor trinomials with a = 1.
- Factor trinomials with a > 1 using AC/grouping.
- Take out the GCF before continuing.
- Recognize a difference of squares.
- Recognize a sum or difference of cubes.
- Factor completely rather than stopping early.
- Use long division and synthetic division appropriately.
Fresh mixed review
| # | Problem | Primary skill |
|---|---|---|
| 1 | 5x²(2x³ − 3x + 4) | Monomial × polynomial |
| 2 | (3x + 4)(5x − 2) | Binomial × binomial |
| 3 | (2x − 5)(x² + 3x − 1) | Binomial × trinomial |
| 4 | x² + 13x + 36 | Factor, a = 1 |
| 5 | x² − 2x − 35 | Factor, a = 1 |
| 6 | 6x² + 13x + 6 | AC method |
| 7 | 12x² − 17x − 5 | AC method |
| 8 | 6x² + 30x + 36 | GCF first, then factor |
| 9 | 64x² − 81 | Difference of squares |
| 10 | 16a² − 49b² | Difference of squares |
| 11 | x³ + 125 | Sum of cubes |
| 12 | 27x³ − 8 | Difference of cubes |
| 13 | (x² + 8x + 15) ÷ (x + 3) | Long division |
| 14 | (2x³ + x² − 13x + 6) ÷ (x − 2) | Synthetic division |
Before you turn in the assessment
- Every multiplication term was distributed; no term was skipped.
- Like terms were combined only when variable parts and exponents matched.
- Every factoring problem began with a GCF check.
- Factored answers were multiplied mentally or on paper to verify the original expression.
- Difference of squares was used only for subtraction of two squares.
- Cube formulas used the correct sign pattern.
- Polynomial dividends were written in descending powers, with zero placeholders for missing terms.
- Synthetic division used the zero of the divisor, not the visible constant with the wrong sign.
Performance Task: Build Polynomial Farm
Farmer Frank's field plan is labeled with polynomials, and your team is going to build it for real. Each team gets a value for x and y. You write the polynomial first, then turn it into feet, stake the plots out with measuring tapes and landscaping flags, and check that what you measure matches what the algebra predicted. The polynomial works the same way for every team. Only the numbers you plug in are different.
Materials (per team of 3–4)
- Two 25–50 ft measuring tapes
- About 20 landscaping flags in 5 colors, one color per crop (e.g., yellow = squash, orange = pumpkins, green = corn, blue = beans, red = potatoes)
- A few flags in a 6th color for the extra plots in Part 3
- String or twine (optional, to mark plot edges)
- A clipboard, the recording sheet, and a pencil
- A team card with the team's x and y values
- A carpenter's square or a 3-4-5 string to get square corners (optional)
- An open space about 25 ft × 15 ft per team (grass, a blacktop, or a gym floor with tape instead of flags)
Farmer Frank's field plan
All lengths are in feet. The plots fit together into one rectangle, so every side of the field has to add up correctly. That gives your team a built-in way to check its work.
Team cards
| Team | x | y | Whole field (W × L) | What to watch for |
|---|---|---|---|---|
| A | 3 | 2 | 17 ft × 10 ft | Teams A and B share the same x, so their west and middle plots should match exactly. |
| B | 3 | 4 | 19 ft × 10 ft | Only the east plots change. Why does y show up in the beans and potatoes but nowhere else? |
| C | 4 | 3 | 21 ft × 12 ft | Compare your squash with Team A's. How much bigger is it, and does it grow at the same rate as the pumpkins? |
| D | 5 | 2 | 23 ft × 14 ft | This is the biggest field. The corn and pumpkins are almost the same area. Is that a coincidence? |
To make your own cards, keep x ≥ 3 (Part 3 explains why) and keep the field size within your space. Whole field = (3x + y + 6) by (2x + 4).
Part 1 — Plan on paper first (before touching a tape)
- Write a polynomial for each labeled side and simplify: the south side of the field and the west side of the field.
- Write and simplify the perimeter of the pumpkin plot.
- Write and simplify the area of every plot: squash, pumpkins, corn, beans, and potatoes. The beans and potatoes will have both x and y in them.
- Add the five plot areas together. Then multiply the whole field's length by its width. These two polynomials have to be equal. If they aren't, find the mistake before you go outside.
- Plug in your team's x and y to get every side length in feet. Your teacher checks the list before you head out.
Part 2 — Stake out the farm
- Start at the northwest corner. Put in the first flag. Stretch a tape east along the north edge and flag each place where one column ends and the next begins.
- Square the corners. Run the west edge south from the first flag at a right angle, using a carpenter's square or the 3-4-5 trick: mark 3 ft on one edge and 4 ft on the other, and adjust until the diagonal between the marks is exactly 5 ft.
- Flag every plot corner in that crop's color. A corner shared by two plots gets both colors. There are 12 corners in all.
- Measure it as built. Measure the whole south side and the pumpkin perimeter with a tape (don't use your planned numbers). Record the measured values next to what you predicted.
- Explain the difference. If a measurement is off by more than about 3 inches, figure out whether the tape, the flag placement, or the algebra caused it.
Part 3 — New plots from area alone
Farmer Frank wants two more plots. He only knows how big each one is (its area), not how long its sides are. Factor to find possible side lengths, plug in your x, then stake out each plot in the 6th flag color next to your farm.
Strawberries: A = 16x² + 4x
Factor out the GCF to get one pair of side lengths. Then find a second, different pair of whole-number side lengths that gives the same area. Stake out whichever pair fits your space better, and measure to confirm the area.
Cucumbers: A = 2x² − x − 6
Factor with the AC method. Before you stake it out, figure out the smallest whole-number x that gives a real plot, and explain what happens to the plot when x = 2.
Part 4 — Double everything
- Pick your squash plot. Write polynomials for a new plot where every side is doubled.
- Stake out the doubled plot on the ground. Then walk its perimeter with the tape and compare it with the original squash perimeter.
- Use string or extra flags to split the doubled plot into copies of the original squash plot. How many copies fit?
- Write one sentence each for what happens to the perimeter and to the area when every side doubles, and back it up with the algebra.
Extension — The mystery plot
Before class, your teacher stakes out one pumpkin plot (2x + 1 by x + 4) using a secret value of x. Measure it and work backward to find x. Then use the other side to check your answer. After that, say what the squash plot next to it would measure.
Recording sheet
| Quantity | Polynomial (simplified) | Predicted (ft or ft²) | Measured | Difference / reason |
|---|---|---|---|---|
| South side of field | ||||
| West side of field | ||||
| Pumpkin perimeter | ||||
| Squash / Pumpkin / Corn areas | — | |||
| Beans / Potato areas | — | |||
| Sum of plots vs. whole field | ||||
| Strawberry dimensions | ||||
| Cucumber dimensions | ||||
| Doubled squash: perimeter & area |
Measured area = your measured length × your measured width. The two tapes let you measure both sides at the same time.
Scoring (/10)
| Component | Points | Full credit looks like |
|---|---|---|
| Side lengths & perimeter | 2 | South side, west side, and pumpkin perimeter are written, simplified, and evaluated correctly. |
| Area polynomials | 2 | All five plot areas are multiplied and simplified, including the two-variable beans and potatoes. |
| Consistency check | 1 | The five plot areas add up to the whole field's area, and the work shows it. |
| Accurate build | 2 | Flags are within about 3 inches of the plan, corners are square, and measured values are recorded and explained. |
| Factoring for new plots | 2 | Both areas are factored, the strawberries have two valid pairs of sides, and the cucumber's limit on x is explained. |
| Doubling conclusion | 1 | Explains that the perimeter doubles and the area is multiplied by 4, using the algebra and the tiled plot as evidence. |
Teacher answer key
Polynomials
- South (and north) side: (2x + 1) + (x + 2) + (y + 3) = 3x + y + 6. West (and east) side: x + (x + 4) = 2x + 4.
- Pumpkin perimeter: 2(2x + 1) + 2(x + 4) = 6x + 10.
- Squash (2x + 1)(x) = 2x² + x. Pumpkins (2x + 1)(x + 4) = 2x² + 9x + 4. Corn (x + 2)(2x + 4) = 2x² + 8x + 8 = 2(x + 2)².
- Beans (y + 3)(x + 1) = xy + 3x + y + 3. Potatoes (y + 3)(x + 3) = xy + 3x + 3y + 9.
- Whole field: (3x + y + 6)(2x + 4) = 6x² + 2xy + 24x + 4y + 24, which is the same as the sum of the five plots. Whole-field perimeter = 10x + 2y + 20.
- Strawberries: 4x(4x + 1). Other valid pairs: 2x(8x + 2) or x(16x + 4). Cucumbers: (2x + 3)(x − 2). At x = 2 the width is 0, so x must be greater than 2.
- Doubled squash: (4x + 2)(2x) = 8x² + 4x = 4(2x² + x). The perimeter doubles, the area is multiplied by 4, and four original squash plots fit inside.
Numbers by team (ft and ft²)
| Team (x, y) | South × West | Pumpkin P | Squash | Pumpkins | Corn | Beans | Potatoes | Field area | Strawberry | Cucumber |
|---|---|---|---|---|---|---|---|---|---|---|
| A (3, 2) | 17 × 10 | 28 | 7×3 = 21 | 7×7 = 49 | 5×10 = 50 | 5×4 = 20 | 5×6 = 30 | 170 | 12×13 = 156 | 9×1 = 9 |
| B (3, 4) | 19 × 10 | 28 | 7×3 = 21 | 7×7 = 49 | 5×10 = 50 | 7×4 = 28 | 7×6 = 42 | 190 | 12×13 = 156 | 9×1 = 9 |
| C (4, 3) | 21 × 12 | 34 | 9×4 = 36 | 9×8 = 72 | 6×12 = 72 | 6×5 = 30 | 6×7 = 42 | 252 | 16×17 = 272 | 11×2 = 22 |
| D (5, 2) | 23 × 14 | 40 | 11×5 = 55 | 11×9 = 99 | 7×14 = 98 | 5×6 = 30 | 5×8 = 40 | 322 | 20×21 = 420 | 13×3 = 39 |
For big strawberry plots (Team D), the 2x by (8x + 2) pair, 10 ft × 42 ft, may fit a long, narrow space better. Mystery plot suggestion: x = 6 gives a 13 ft × 10 ft pumpkin plot, and the squash next to it is 13 ft × 6 ft.
Chapter 3 Basis
This page follows the sequence and methods in the provided Chapter 3 notes and uses the attached review and performance task to determine the assessment emphasis. Practice questions on this page are newly written analogues rather than copies of the review sheet.