Polynomials path · A08

Before this lesson: Polynomial division, Solving polynomial equations

Your goal: Reduce the degree of an equation satisfied by a shared root.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Common roots of polynomials: the key idea

If a number r is a common root of P and Q, it is a root of every polynomial combination AP+BQ. Subtract suitable multiples to cancel leading terms and reduce the degree. This is the polynomial version of the Euclidean algorithm. A common factor records shared roots over the chosen field. Parameter problems need care: a coefficient used for division may be zero. Solve the lower-degree condition, substitute back into both original equations, and separate every exceptional parameter case.

A worked example

Find the common roots of x²−5x+6 and x²−4x+3.

Subtracting the second polynomial from the first gives −x+3. Any common root must therefore be x=3. Both originals vanish at 3, so it is a common root. Since the necessary linear condition has only one root, the list is complete.

Your turn: change one thing

For which real k do x²−3x+2 and x²+kx−2 have a common root?

Try this on paper before opening the explanation.

Compare your reasoning

The first polynomial factors as (x−1)(x−2). Substituting 1 into the second gives k−1=0, so k=1. Substituting 2 gives 2+2k=0, so k=−1. Each value gives the indicated common root, hence k∈{1,−1}.

Pause and check

A trap to avoid: Dividing by a parameter before separating its zero case.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Prove that P and Q have exactly the same common roots as Q and P−AQ, for any polynomial A.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

If P(r)=Q(r)=0, then P(r)−A(r)Q(r)=0. Conversely, if Q(r)=0 and P(r)−A(r)Q(r)=0, substituting the first equality into the second gives P(r)=0. Both directions hold, so the sets of common roots agree.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. If r is a common root of P and Q, which expression must vanish at r?

Foundation

  1. P·Q+1
  2. P−Q
  3. P+1
  4. Q+1
Hint

Evaluate using both zeros.

Answer and reasoning

P−Q. P(r)−Q(r)=0−0=0.

2. Which polynomial shares root 2 with x−2?

Foundation

  1. x+2
  2. x²−2
  3. x²−4
  4. x²+4
Hint

Substitute 2.

Answer and reasoning

x²−4. 2²−4=0, while the other values are nonzero.

3. Find the common root of x²−5x+6 and x²−4x+3.

Core

  1. 1
  2. 2
  3. 4
  4. 3
Hint

Subtract the equations.

Answer and reasoning

3. A common root satisfies −x+3=0. Checking x=3 in both confirms it.

4. Find the monic gcd of (x−1)(x−2) and (x−2)(x−3).

Core

  1. x−2
  2. x−1
  3. x−3
  4. (x−1)(x−3)
Hint

Find the factor present in both.

Answer and reasoning

x−2. The only common linear factor is x−2, with multiplicity one.

5. After eliminating a term, why substitute candidates into both originals?

Stretch

  1. A common root cannot be real
  2. Every candidate is automatically valid
  3. Elimination may give necessary but insufficient conditions
  4. Substitution changes the roots
Hint

A combination can have extra roots.

Answer and reasoning

Elimination may give necessary but insufficient conditions. A root of P−Q need not be a root of either P or Q. Checking restores the original requirements.

6. An elimination step divides by k. What must be done?

Stretch

  1. Assume k is positive
  2. Replace k with 1
  3. Ignore that case
  4. Handle k=0 separately
Hint

A division may remove an allowed parameter case.

Answer and reasoning

Handle k=0 separately. The transformation is reversible only for k≠0; the original equations must settle k=0.

7. If r is a common root of P and Q, what is P(r)−Q(r)?

Foundation

  1. 0
  2. 1
  3. r
  4. P(0)
Hint

Both polynomial values are zero.

Answer and reasoning

0. 0−0=0.

8. Find the common root of x²−3x+2 and x²−5x+6.

Core

  1. −2
  2. 2
  3. 1
  4. 3
Hint

Factor both quadratics.

Answer and reasoning

2. Their root sets are {1,2} and {2,3}, whose intersection is {2}.

9. Why must a candidate from P−Q=0 be checked in P=0?

Stretch

  1. The difference may have roots that neither original has
  2. Subtraction always loses all roots
  3. Only Q matters
  4. Polynomials have infinitely many common roots
Hint

A necessary condition need not be sufficient.

Answer and reasoning

The difference may have roots that neither original has. P(r)=Q(r) can hold at a nonzero common value.

Choose your next step

Continue to Irreducible polynomials. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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