Digital SAT Math · Advanced Math
Polynomial operations and structure
Most polynomial products on the Digital SAT do not ask for the whole expansion. They ask for one coefficient, or the degree, or what the graph does at the ends — and those answers come from a handful of multiplications, not from writing out every term. The skill is knowing which multiplications matter and leaving the rest on the table.
On the test
| Domain | Advanced Math (score report) |
| What it looks like | Two or three polynomials to add, subtract, or multiply; a request for one coefficient, the degree of a product, or the end behaviour of |
| Often asked | “What is the coefficient of …?”, “What is the value of …?”, “What is the degree of…?”, “Which statement describes the end behaviour…?” |
| Format | Multiple choice and student-produced response |
| Calculator | Allowed; almost never faster than the target-degree method for clean coefficients — see the box below |
Recognition cues: equivalent to, coefficient of, leading coefficient, degree of the product, as approaches, for all values of when matching a partial expansion, parentheses stacked for a product of three factors.
What this page owns
College Board files “add, subtract, and multiply polynomials” next to factoring under one skill label, and free guides copy that lump. We split it. Here the work is arithmetic and structure: combine like terms, distribute a minus, multiply, read degree and leading coefficient, predict end behaviour. When the rewrite is the answer — factoring, difference of squares, cancelling in a rational — that is equivalent expressions. When a factored form is a step toward a root, that is nonlinear equations. When the point of a form is a vertex or intercept on a parabola, that is quadratic functions.
Pattern recognition
Five shapes cover essentially every item:
- Full expand — choices are complete polynomials; every term of the product is needed.
- Single coefficient — the stem asks for the coefficient of only; only pairs whose degrees sum to matter.
- Add / subtract — line up like terms; a minus distributes to every term of the second polynomial.
- Degree of a product (or sum) — degrees add under multiplication when leading coefficients are nonzero; leading terms can cancel under addition or subtraction, so simplify before reading degree.
- End behaviour — parity of the degree plus the sign of the leading coefficient fix both ends of the graph.
Method
- Name the ask. One coefficient, a full polynomial, a degree, or an end-behaviour statement. Say it in five words before multiplying anything.
- If the ask is one coefficient of degree : list every pair of terms (one from each factor) whose degrees sum to . Multiply coefficient × coefficient for each pair, add those products. That sum is the answer. Ignore every other pair.
- If the ask is a full polynomial: distribute every term of the first factor across the second (or third), then combine like terms. Carry every sign — a minus in front of a parenthesis flips all terms inside it.
- If the ask is degree: for a product of nonzero polynomials, add the degrees. For a sum or difference, simplify first — leading terms may cancel and drop the degree.
- If the ask is end behaviour: read only the leading term . Even → both ends the same direction; odd → opposite ends. Positive → right end up; negative → right end down.
- Check with one number when you expanded fully: pick (avoid and ) and compare the original product to your result. On a single-coefficient item the check is re-listing the pairs.
| Stem asks for | What you compute |
|---|---|
| coefficient of | only pairs with degrees summing to |
| full expanded form | every product, then combine |
| degree of | (leading coeffs nonzero) |
| degree of | simplify first, then highest remaining power |
| end behaviour | sign of leading coefficient × parity of degree |
| leading coefficient of a product | product of the two leading coefficients |
Worked example 1 — one coefficient, not the whole product
Stem. The expression is equivalent to , where and are constants. What is the value of ?
Step 1 — name the ask. Coefficient of , not the full expansion.
Step 2 — pairs that make degree 2.
- from the first factor with the constant from the second:
- from the first with from the second:
- The constant would need an from the second factor; there is none.
Step 3 — add those contributions.
Check (optional full expand). . The coefficient is 6, and the constant 12 matches the stem. Answer: .
Trap watch. Reporting uses only the first pair. Reporting uses only the second. Reporting is the leading coefficient of the product, which is the coefficient of , not of .
Worked example 2 — distribute the minus, then combine
Stem. Which expression is equivalent to ?
Step 1 — distribute the minus to every term of the second polynomial.
The becomes ; the becomes . Both flips are where students lose the item.
Step 2 — combine like terms.
Check. At : original left side is ; result is ✓. Answer: .
Trap watch. Leaving unflipped produces . Forgetting the final produces a constant of 6 instead of 8. Adding the two polynomials instead of subtracting produces a leading term .
Worked example 3 — end behaviour from two facts
Stem. For the polynomial function , which statement is true about the end behaviour of the graph of in the -plane?
Step 1 — read only the leading term. . Degree (even). Leading coefficient (negative).
Step 2 — apply the two-bit rule.
- Even degree → both ends go the same direction.
- Negative leading coefficient → that direction is down.
So as , , and as , .
Step 3 — discard the other three patterns. Odd degree with positive leading would rise to the right and fall to the left; even degree with positive leading would rise on both ends; odd degree with negative leading would fall to the right and rise to the left. None of those matches even + negative.
Check. For large , dominates every lower term; is large and negative, and the same is true for . Answer: both ends go to .
Trap watch. Reading the constant as if it controlled the ends. Reading the degree as odd because of the term sitting next to the leading term. Flipping only one end.
Practice
Answer first, then open the explanation. Two items are student-produced response (type the number, no choices). Every wrong choice is one named arithmetic or structure error — a flipped sign, a degree product instead of a degree sum, a missing cross term, a leading coefficient treated as a constant. Log the error type, not the item number.
Question 1 Warm-up
Which of the following expressions is equivalent to (4x − 3)(x + 2)?
Show the answer Choice B
Why it is right
Multiply every term of the first binomial by every term of the second: 4x·x = 4x², 4x·2 = 8x, −3·x = −3x, and −3·2 = −6. The two middle products combine as 8x − 3x = 5x, so the expansion is 4x² + 5x − 6. Checking at x = 1: the original product is (4 − 3)(1 + 2) = 3, and 4 + 5 − 6 = 3.
Why each other choice fails
- Choice A
- Gets the middle term right but multiplies −3·2 as +6 instead of −6. A negative times a positive is negative, so the constant term of the product is −6.
- Choice C
- Flips the sign of the combined middle term, writing −5x instead of +5x. The products 8x and −3x combine to +5x, not −5x.
- Choice D
- Keeps only the outer middle product 8x and drops the inner product −3x. Both cross products contribute to the coefficient of x, so 8 + (−3) = 5.
Question 2 Standard
The expression (3x + 4)(2x − 5) is equivalent to 6x² + bx − 20, where b is a constant. What is the value of b?
Show the answer Choice C
Why it is right
Only the two cross products can produce an x-term: 3x·(−5) = −15x and 4·2x = 8x. Adding those contributions gives (−15 + 8)x = −7x, so b = −7. The leading term 3x·2x = 6x² and the constant 4·(−5) = −20 already match the stem, which confirms the factors were read correctly without a full rewrite of every power.
Why each other choice fails
- Choice A
- Adds the sizes of the two cross products and ignores that one is negative: 15 + 8 = 23. Only 8x is positive; the other product is −15x, so they partly cancel to −7.
- Choice B
- Reports only the outer cross product 3x·(−5) = −15 and never adds the inner product 4·2x = 8x. Both pairs contribute to the coefficient of x.
- Choice D
- Reports only the inner cross product 4·2x = 8x and drops the outer product −15x. The coefficient of x is the sum of both, −15 + 8 = −7.
Question 3 Standard
Which of the following expressions is equivalent to (5x³ − 2x² + 7) − (2x³ − 4x² + 3x − 1)?
Show the answer Choice A
Why it is right
Distribute the minus sign to every term of the second polynomial: 5x³ − 2x² + 7 − 2x³ + 4x² − 3x + 1. The −4x² flips to +4x² and the −1 flips to +1. Combining like terms then gives (5 − 2)x³ + (−2 + 4)x² − 3x + (7 + 1) = 3x³ + 2x² − 3x + 8. At x = 1 both the original difference and this result equal 10.
Why each other choice fails
- Choice B
- Distributes the minus to the leading 2x³ but leaves −4x² unflipped, so the x² coefficient becomes −2 − 4 = −6 instead of −2 + 4 = 2. Every term inside the subtracted parentheses changes sign.
- Choice C
- Handles the variable terms correctly but forgets that subtracting −1 adds 1 to the constant, so the constant is written as 7 − 1 = 6 instead of 7 + 1 = 8.
- Choice D
- Adds the two leading coefficients, 5 + 2 = 7, instead of subtracting them. The operation between the polynomials is subtraction, so the x³ coefficient is 5 − 2 = 3.
Question 4 Standard
The polynomial f has degree 3 and the polynomial g has degree 2. Neither leading coefficient is zero. What is the degree of the product f(x)·g(x)?
Show the answer Choice D
Why it is right
When two nonzero polynomials are multiplied, the degree of the product equals the sum of the degrees, provided neither leading coefficient is zero. Here deg f = 3 and deg g = 2, so deg(f·g) = 3 + 2 = 5. The leading term of the product is the product of the two leading terms, and that single term has degree 5, so no higher or lower degree can appear as the leading degree.
Why each other choice fails
- Choice A
- Multiplies the two degrees, 3 · 2 = 6, instead of adding them. Degree multiplies with the variable when a single term is powered, not when two polynomials are multiplied.
- Choice B
- Reports the larger of the two degrees. That rule is closer to what happens with addition when leading terms do not cancel; for a product the degrees add.
- Choice C
- Subtracts the degrees, 3 − 2 = 1. Subtraction of degrees would describe a quotient of monic leading terms, not a product.
Question 5 Standard
Which of the following expressions is equivalent to (2x − 7)²?
Show the answer Choice A
Why it is right
A squared binomial is a product: (2x − 7)² = (2x − 7)(2x − 7). Expanding gives (2x)(2x) = 4x², (2x)(−7) = −14x, (−7)(2x) = −14x, and (−7)(−7) = 49. The two identical cross products combine into −28x, so the square is 4x² − 28x + 49 — the pattern (a − b)² = a² − 2ab + b² with a = 2x and b = 7.
Why each other choice fails
- Choice B
- Squares the two terms separately, as if (a − b)² were a² + b². That discards both cross products; at x = 1 the original is (−5)² = 25 while 4 + 49 = 53.
- Choice C
- Counts the cross product once instead of twice, writing −14x for the middle term. Multiplying out produces −14x from 2x·(−7) and another −14x from (−7)·2x, so the middle term is −28x.
- Choice D
- Gets the middle term right but multiplies (−7)(−7) as −49 instead of +49. Two negatives multiply to a positive, so the constant term of a square is always positive.
Question 6 Harder Student-produced response
The expression (x² + 5x − 3)(2x − 4) is equivalent to 2x³ + ax² − 26x + 12, where a is a constant. What is the value of a?
Show the answer 6
Why it is right
Only pairs whose degrees sum to 2 contribute to the coefficient of x². The pairs are x² with −4, giving 1 · (−4) = −4, and 5x with 2x, giving 5 · 2 = 10. The constant −3 has no x² partner in the second factor. Adding the contributions yields a = −4 + 10 = 6. A full expansion, 2x³ + 6x² − 26x + 12, confirms both the asked coefficient and the constant 12 printed in the stem.
Answers students type instead
- 2
- Reports the leading coefficient of the product, which is the coefficient of x³ from x² · 2x, not the coefficient of x² the question asked for.
- 10
- Uses only the pair 5x · 2x and drops x² · (−4) = −4. The coefficient of x² is the sum −4 + 10 = 6, not either contribution alone.
- -4
- Uses only the pair x² · (−4) and never adds the second pair 5x · 2x = 10. Both pairs produce degree 2, so both belong in a.
Question 7 Harder
For the polynomial function f(x) = −2x⁴ + 5x³ − x + 8, which of the following statements about the end behaviour of the graph of y = f(x) in the xy-plane is true?
Show the answer Choice B
Why it is right
End behaviour is fixed by the leading term alone, here −2x⁴. The degree 4 is even, so both ends of the graph go in the same direction. The leading coefficient −2 is negative, so that shared direction is down: as x → ∞, f(x) → −∞, and as x → −∞, f(x) → −∞. For a concrete check, f(10) is dominated by −2·10000 = −20000 and f(−10) is dominated by the same −20000, both large and negative.
Why each other choice fails
- Choice A
- Describes the end behaviour of an odd-degree polynomial with positive leading coefficient (right end up, left end down). The degree here is 4, which is even, so the two ends cannot go opposite ways.
- Choice C
- Describes an even-degree polynomial with positive leading coefficient (both ends up). The leading coefficient here is −2, not positive, so both ends go down rather than up.
- Choice D
- Describes an odd-degree polynomial with negative leading coefficient (right end down, left end up). Degree 4 is even, so the ends match each other; they do not oppose.
Question 8 Harder
The product (x + 6)(x² − 2x + 3) is written as x³ + px² + qx + 18, where p and q are constants. What is the value of p?
Show the answer Choice C
Why it is right
The coefficient of x² comes only from pairs whose degrees sum to 2: x · (−2x) = −2x² and 6 · x² = 6x². Adding those contributions gives p = −2 + 6 = 4. The constant term 6 · 3 = 18 already matches the stem, and the leading term x · x² = x³ matches the printed x³, so the factors were read correctly.
Why each other choice fails
- Choice A
- Reports only the contribution x · (−2x) = −2 and never adds 6 · x² = 6. Both pairs produce degree 2, so p is their sum, 4.
- Choice B
- Reports only the contribution 6 · x² = 6 and drops x · (−2x) = −2. The coefficient of x² is −2 + 6 = 4, not 6 alone.
- Choice D
- Subtracts the two contributions instead of adding them, computing −2 − 6 = −8. Like-term coefficients combine by addition: −2x² + 6x² = 4x².
Question 9 Harder
Which of the following expressions is equivalent to (4x³ − x + 2) − (2x³ + 3x² − 5) + (x³ − x² + 4x)?
Show the answer Choice D
Why it is right
Distribute the minus through the middle polynomial and then combine: 4x³ − x + 2 − 2x³ − 3x² + 5 + x³ − x² + 4x. The x³ terms give 4 − 2 + 1 = 3; the x² terms give −3 − 1 = −4; the x terms give −1 + 4 = 3; the constants give 2 + 5 = 7. The result is 3x³ − 4x² + 3x + 7. At x = 1 the original expression and this result both equal 9.
Why each other choice fails
- Choice A
- Distributes the minus to 3x² but then combines the x² terms as −3 + 1 instead of −3 − 1, as if the third polynomial contributed +x². The third polynomial contributes −x², so the x² coefficient is −4.
- Choice B
- Handles the variable terms correctly but treats the constants as 2 − 5 = −3 instead of 2 − (−5) = 2 + 5 = 7. Subtracting −5 adds 5.
- Choice C
- Adds the middle polynomial's leading term instead of subtracting it, computing 4 + 2 + 1 = 7 for x³ or similar, and lands on 5x³. The middle 2x³ is subtracted, so the x³ coefficient is 4 − 2 + 1 = 3.
Question 10 Hardest
The product (x − 1)(x + 2)(2x − 3) is equivalent to 2x³ + ax² + bx + 6, where a and b are constants. What is the value of a?
Show the answer Choice B
Why it is right
First multiply two factors: (x − 1)(x + 2) = x² + x − 2. Then multiply by the third: (x² + x − 2)(2x − 3) = 2x³ − 3x² + 2x² − 3x − 4x + 6 = 2x³ − x² − 7x + 6. The coefficient of x² is −1, so a = −1. The constant (−2)(−3) = 6 and the leading 1 · 1 · 2 = 2 match the stem, confirming the expansion.
Why each other choice fails
- Choice A
- Flips the sign of the x² coefficient, reporting +1 instead of −1. The contributions to x² are −3 from x² · (−3) and +2 from x · 2x, which sum to −1, not +1.
- Choice C
- Uses a wrong middle for the first product — writing (x − 1)(x + 2) as x² − x − 2 instead of x² + x − 2 — and then carries that error into the x² coefficient of the full product.
- Choice D
- Adds the constants from two of the factors, or otherwise combines numbers that never produce an x² term. The x² coefficient is built only from pairs whose degrees sum to 2 inside the stepwise product.
Question 11 Hardest Student-produced response
The product (3x − 1)(x + 4)(x − 2) is equivalent to 3x³ + 5x² + cx + 8, where c is a constant. What is the value of c?
Show the answer -26
Why it is right
First multiply the last two factors: (x + 4)(x − 2) = x² + 2x − 8. Then multiply by (3x − 1): 3x(x² + 2x − 8) − 1(x² + 2x − 8) = 3x³ + 6x² − 24x − x² − 2x + 8. Combining like terms gives 3x³ + 5x² − 26x + 8, so c = −26. The printed leading coefficient 3 and constant 8 match 3 · 1 · 1 and (−1)(4)(−2) = 8, which confirms the factors.
Answers students type instead
- 26
- Drops the overall minus on the x-coefficient after combining −24x − 2x correctly as magnitude 26. The x-terms are both negative, so c = −26, not 26.
- -24
- Keeps only the contribution 3x · (−8) = −24x from the first row and never adds (−1) · (2x) = −2x. Both rows contribute to the coefficient of x.
- -22
- Adds −24x to +2x instead of −2x, as if the second row contributed a positive middle term. Distributing −1 flips the sign of 2x to −2x, so −24 + (−2) = −26.
Question 12 Hardest
The polynomial p is defined by p(x) = (2x² − 5)(x³ + 3x − 1) − (x⁵ − 4x³ + 7). What is the leading coefficient of p?
Show the answer Choice A
Why it is right
The highest degree in the product (2x² − 5)(x³ + 3x − 1) comes from 2x² · x³ = 2x⁵, so that product is 2x⁵ + (lower terms). Subtracting x⁵ − 4x³ + 7 removes one x⁵: 2x⁵ − x⁵ = x⁵. No other term in either piece has degree greater than 5, so the leading term of p is 1 · x⁵ and the leading coefficient is 1. Fully expanded, p(x) = x⁵ + 5x³ − 2x² − 15x − 2, which confirms the leading coefficient without changing it.
Why each other choice fails
- Choice B
- Stops after reading the leading coefficient of the product, 2 · 1 = 2, and never subtracts the leading term of the second polynomial. The subtraction 2x⁵ − x⁵ changes the leading coefficient from 2 to 1.
- Choice C
- Subtracts in the wrong order or flips the sign of the product's leading term, treating the x⁵ contribution as −2x⁵ + x⁵ = −x⁵. The product contributes +2x⁵ and the subtracted polynomial contributes −x⁵, leaving +x⁵.
- Choice D
- Assumes the two x⁵ terms cancel completely. They would cancel only if the product's leading coefficient were also 1; it is 2, so one x⁵ remains.
Common mistakes
- Expanding everything when only one coefficient is asked — writing all six or nine products and then mis-adding, when two pairs would have been enough.
- Sign lost on a distributed minus — subtracting as if only the changed sign.
- — discarding the cross term ; the middle term is the whole point of the pattern.
- Counting the cross term once in a square — written with middle term instead of .
- Degrees multiplied instead of added — claiming a degree-3 times a degree-2 product has degree 6.
- Degree read before simplifying a sum or difference — leading terms cancel and the true degree drops.
- Leading coefficient confused with the constant — using the trailing number to predict end behaviour.
- Even/odd parity flipped — treating an even-degree polynomial as if its ends went opposite ways.
- Only one pair collected for a middle coefficient — the -term of a binomial product comes from two cross products, not one.
- Two negatives multiplying to a negative — written as in a constant term.
- Answering the leading coefficient when asked for the coefficient of — different powers, different pairs.
FAQ
Do I always need to FOIL? No. FOIL is full expansion of two binomials. Use it when the choices are full polynomials. When the stem asks for one coefficient, list only the pairs that produce that degree — often two multiplications and an add.
How do I find the coefficient of in a product of three factors? Either multiply two factors fully and then run the target-degree method against the third, or systematically list every triple of terms whose degrees sum to . The two-then-one route is less error-prone under time pressure.
What is the degree of a product? If neither factor is the zero polynomial and neither leading coefficient is zero, . The leading coefficient of the product is the product of the leading coefficients.
How do I predict end behaviour without graphing? Look only at . Even : both ends same way (up if , down if ). Odd : opposite ends (right end up if , right end down if ). Lower-degree terms cannot change either end.
Is factoring on this page? No. Factoring to expose structure or roots is equivalent expressions and nonlinear equations. If you catch yourself factoring to answer a coefficient question, you are on the slow path — multiply the pairs instead.
Grid-in tips? Enter the exact integer or fraction. Negative coefficients are common; a leading minus is allowed. If the question asks for in , enter only — not the whole trinomial and not .