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The polynomial has equal remainders on division by x−2 and x+1. Find a.

Read the idea, work independently, then explain what changed.

TOPIC 01

2022 JM01

The root of x+1 is −1, not 1.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

The polynomial has equal remainders on division by x−2 and x+1. Find a.

f(x)=x3+3x2+ax+bf(x)=x^3+3x^2+ax+b

Official paper · jm01-2022 · I.3 · PDF 2

Official original and suggested answers ↗ · Suggested answer PDF page 5

Skills and prerequisite lessons
  1. Option A44
  2. Option B−3-3
  3. Option C−4-4
  4. Option D99
  5. Option E−6-6

Working and explanation

BUILD THE REASONING

Hint 1
Use f(2)=f(−1).
Hint 2
The constant b cancels.
Worked solution
  1. Evaluate the two remainders.

    f(2)=20+2a+b,f(−1)=2−a+bf(2)=20+2a+b,\quad f(-1)=2-a+b
  2. Solve the resulting linear equation.

    20+2a=2−a  ⟹  a=−620+2a=2-a\implies a=-6

E: −6.

Checks and common pitfalls: The root of x+1 is −1, not 1.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Evaluate the two remainders.
    f(2)=20+2a+b,f(−1)=2−a+bf(2)=20+2a+b,\quad f(-1)=2-a+b
  • Solve the resulting linear equation.
    20+2a=2−a  ⟹  a=−620+2a=2-a\implies a=-6

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Curriculum and source notes ↗