Polynomial factors and multiples · Teacher reference · answers included

← Senior 1 · short mathematical inquiries

Teacher reference: contains answers, conditions and proofs.

12 Polynomial factors and multiples · g10-12-01 · 5–10 min · TEACHER · ANSWERS

Which factor copies are shared?

Why does a greatest common divisor use the lower exponent?

Predict: Build A=(x−1)ᵐ(x+1)ⁿ and B=(x−1)ᵖ(x+1)ᑫ; predict their shared tiles.

Compare	Computed observation
Two shared factors A: exponent of x−1 = 2; A: exponent of x+1 = 1; B: exponent of x−1 = 1; B: exponent of x+1 = 3	Monic result: GCD = (x−1)^1(x+1)^1
Coprime factors A: exponent of x−1 = 2; A: exponent of x+1 = 0; B: exponent of x−1 = 0; B: exponent of x+1 = 2	Monic result: GCD = (x−1)^0(x+1)^0
One polynomial is one A: exponent of x−1 = 0; A: exponent of x+1 = 0; B: exponent of x−1 = 2; B: exponent of x+1 = 1	Monic result: GCD = (x−1)^0(x+1)^0

Conclusion: A common divisor cannot use more copies than either polynomial contains, so it uses the minimum exponent of each factor. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: Both polynomials are monic and nonzero. Exponent zero contributes the factor 1.

Justification: For each distinct irreducible factor, a common divisor exponent is at most both source exponents, hence at most their minimum. The constructed product itself divides both.

Ask: Request a divisibility check by subtracting factor counts.

Watch for: Greatest refers to divisibility, not whichever evaluated number is largest.

Extension: Show that every common divisor divides the displayed greatest common divisor.

Exit: Find the monic GCD of (x−1)²(x+1) and (x−1)(x+1)³.

(x−1)(x+1).

Before: Factorisation and divisibility · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

12 Polynomial factors and multiples · g10-12-02 · 5–10 min · TEACHER · ANSWERS

The smallest shared multiple

Why is multiplying A and B usually excessive?

Predict: Compare all factor copies in AB with the copies needed by a common multiple.

Compare	Computed observation
Shared and unshared copies A: exponent of x−1 = 2; A: exponent of x+1 = 1; B: exponent of x−1 = 1; B: exponent of x+1 = 3	Monic result: LCM = (x−1)^2(x+1)^3
Equal polynomials A: exponent of x−1 = 2; A: exponent of x+1 = 1; B: exponent of x−1 = 2; B: exponent of x+1 = 1	Monic result: LCM = (x−1)^2(x+1)^1
One polynomial is one A: exponent of x−1 = 0; A: exponent of x+1 = 0; B: exponent of x−1 = 2; B: exponent of x+1 = 1	Monic result: LCM = (x−1)^2(x+1)^1

Conclusion: A common multiple must contain enough copies for both source polynomials, so each exponent is the maximum, not the sum. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: All polynomials are monic and nonzero, with only the displayed distinct factors.

Justification: A common-multiple exponent is at least both source exponents, hence at least their maximum. The construction attains it; min(u,v)+max(u,v)=u+v proves the product identity.

Ask: Ask which factor can be removed without losing either divisibility requirement.

Watch for: A multiple need not include two separate copies of a factor already shared.

Extension: Verify GCD(A,B)·LCM(A,B)=AB factor by factor.

Exit: Find the monic LCM of x−1 and (x−1)².

(x−1)², not (x−1)³.

Before: GCD and factor counts · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

12 Polynomial factors and multiples · g10-12-03 · 5–10 min · TEACHER · ANSWERS

Why do remainders preserve the GCD?

Can replacing A by a remainder lose a common factor?

Predict: Start with A=(x−r)(x²+u), B=(x−r)(x+v); predict the final nonzero remainder.

Compare	Computed observation
A linear GCD Shared root r = 1; Constant u in x²+u = 1; Constant v in x+v = 2	Monic GCD: x −1
B divides A Shared root r = 1; Constant u in x²+u = -1; Constant v in x+v = 1	Monic GCD: x^2 −1
A root at zero Shared root r = 0; Constant u in x²+u = 0; Constant v in x+v = 0	Monic GCD: x^2

Conclusion: The last nonzero Euclidean remainder, made monic, is the GCD. Sometimes B already divides A. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: The displayed polynomials are nonzero; each remainder has smaller degree. The zero remainder terminates the algorithm.

Justification: If A=BQ+R, a divisor common to A,B divides R=A−BQ; conversely a divisor of B,R divides A. Repetition preserves all common divisors until termination.

Ask: After every arrow, ask pupils to justify the same-common-divisors claim in both directions.

Watch for: A zero remainder is not itself the GCD; use the preceding nonzero divisor.

Extension: Find a condition on u,v that makes the GCD quadratic rather than linear.

Exit: Why do A and B have the same GCD as B and A−B?

A common divisor of A,B divides A−B; a common divisor of B,A−B divides their sum A.

Before: Polynomial division and factors · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.