AMC 10 · 2025 · #7
Grade 8 algebraSuppose a and b are real numbers. When the polynomial x3+x2+ax+b is divided by x−1, the remainder is 4. When the polynomial is divided by x−2, the remainder is 6. What is b−a?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The cubic $f(x) = x^3 + x^2 + ax + b$ leaves remainder $4$ when divided by $x - 1$ and remainder $6$ when divided by $x - 2$. Find $b - a$.
Givens: $f(x) = x^3 + x^2 + ax + b$ with $a$ and $b$ real; Dividing $f(x)$ by $x - 1$ leaves remainder $4$; Dividing $f(x)$ by $x - 2$ leaves remainder $6$; Answer choices: (A) $14$, (B) $15$, (C) $16$, (D) $17$, (E) $18$
Unknowns: The value of $b - a$
Understand
Restated: The cubic $f(x) = x^3 + x^2 + ax + b$ leaves remainder $4$ when divided by $x - 1$ and remainder $6$ when divided by $x - 2$. Find $b - a$.
Givens: $f(x) = x^3 + x^2 + ax + b$ with $a$ and $b$ real; Dividing $f(x)$ by $x - 1$ leaves remainder $4$; Dividing $f(x)$ by $x - 2$ leaves remainder $6$; Answer choices: (A) $14$, (B) $15$, (C) $16$, (D) $17$, (E) $18$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems
Long division looks like the obvious route, but the Remainder Theorem lets you skip it entirely: dividing $f(x)$ by $x - c$ leaves remainder $f(c)$, so each remainder fact is just the value of $f$ at a point. Tool #7 (Identify Subproblems) sees the two divisions as two independent clues, and Tool #4 (Introduce a Variable) turns each clue into an equation in the unknowns $a$ and $b$. Two equations, two unknowns — solve the pair and read off $b - a$.
Execute — Answer: E
6.EE.A.2 Step 1 Read remainders as values
- The two divisions are two separate facts, so treat each on its own.
- The Remainder Theorem says that dividing $f(x)$ by $x - c$ leaves remainder $f(c)$, because $f(x) = (x - c)\,q(x) + r$ and setting $x = c$ kills the first term.
- So the remainder-$4$ fact becomes $f(1) = 4$, and the remainder-$6$ fact becomes $f(2) = 6$.
💡 Plugging in $x = c$ erases the $(x - c)$ factor, so the leftover is exactly the remainder.
6.EE.A.2 Step 2 Evaluate at x = 1
- Substitute $x = 1$ into $f(x) = x^3 + x^2 + ax + b$.
- The numeric terms give $1 + 1 = 2$, and $ax + b$ becomes $a + b$.
- Setting this equal to $4$ collapses to a clean linear equation in $a$ and $b$.
💡 One point turns the whole cubic into a single straight-line relationship between $a$ and $b$.
6.EE.A.2 Step 3 Evaluate at x = 2
- Substitute $x = 2$.
- Now $2^3 + 2^2 = 8 + 4 = 12$, and $ax + b$ becomes $2a + b$.
- Setting the total equal to $6$ gives a second linear equation.
💡 A second point gives a second line; where the two lines meet is the only $(a, b)$ that fits both.
8.EE.C.8 Step 4 Solve the two equations
- Both equations share the term $b$, so subtracting equation (i) from equation (ii) cancels $b$ and isolates $a$.
- That gives $a = -8$.
- Put $a = -8$ back into (i) to get $b = 10$.
💡 Subtracting the equations wipes out the matching $b$ term, leaving one unknown to solve at once.
7.NS.A.1 Step 5 Compute b − a
- The question asks for $b - a$, not $a$ or $b$ alone.
- With $b = 10$ and $a = -8$, subtracting a negative adds: $b - a = 10 - (-8) = 18$.
- That matches choice (E).
💡 Subtracting a negative number is the same as adding its size, so $10 - (-8) = 18$.
6.EE.A.2 The two divisions are two separate facts, so treat each on its own. The Remainde 6.EE.A.2 Substitute $x = 1$ into $f(x) = x^3 + x^2 + ax + b$. The numeric terms give $1 + 6.EE.A.2 Substitute $x = 2$. Now $2^3 + 2^2 = 8 + 4 = 12$, and $ax + b$ becomes $2a + b$. 8.EE.C.8 Both equations share the term $b$, so subtracting equation (i) from equation (ii 7.NS.A.1 The question asks for $b - a$, not $a$ or $b$ alone. With $b = 10$ and $a = -8$, Review
Reasonableness: Rebuild the polynomial with $a = -8$ and $b = 10$: $f(x) = x^3 + x^2 - 8x + 10$. Then $f(1) = 1 + 1 - 8 + 10 = 4$ and $f(2) = 8 + 4 - 16 + 10 = 6$ — both remainders come out exactly as stated, so the values are correct and $b - a = 18$.
Alternative: Skip solving for $a$ and $b$ separately. Subtracting equation (i) from (ii) gives $a = -8$ directly, and only then is $b = 10$ needed. Or notice the target $b - a$ can be built by combining the two equations: from $a + b = 2$ and $2a + b = -6$, the value $b - a = (a + b) - 2a = 2 - 2(-8) = 18$, reaching the answer with the same two facts.
CCSS standards used (min grade 8)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Evaluating $f(x) = x^3 + x^2 + ax + b$ at $x = 1$ and $x = 2$ to turn each remainder fact into an equation in $a$ and $b$.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving the system $a + b = 2$ and $2a + b = -6$ by elimination to find $a = -8$ and $b = 10$.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Computing $b - a = 10 - (-8) = 18$, subtracting a negative number correctly.)
⭐ A remainder after dividing by $x - c$ is just the value $f(c)$, so plug in each number to build equations instead of doing long division.
⭐ A remainder after dividing by $x - c$ is just the value $f(c)$, so plug in each number to build equations instead of doing long division.
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