12 Polynomial factors and multiples · g10-12-01 · 5–10 min
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.
Case numbers match the projection buttons. Record the displayed conditions and your observation.
| Compare | What changed? What stayed? |
|---|
Case 1
A: exponent of x−1 = 2; A: exponent of x+1 = 1; B: exponent of x−1 = 1; B: exponent of x+1 = 3 | |
Case 2
A: exponent of x−1 = 2; A: exponent of x+1 = 0; B: exponent of x−1 = 0; B: exponent of x+1 = 2 | |
Case 3
A: exponent of x−1 = 0; A: exponent of x+1 = 0; B: exponent of x−1 = 2; B: exponent of x+1 = 1 | |
Revise: My claim needs the condition… because…
Exit: Find the monic GCD of (x−1)²(x+1) and (x−1)(x+1)³.
12 Polynomial factors and multiples · g10-12-02 · 5–10 min
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.
Case numbers match the projection buttons. Record the displayed conditions and your observation.
| Compare | What changed? What stayed? |
|---|
Case 1
A: exponent of x−1 = 2; A: exponent of x+1 = 1; B: exponent of x−1 = 1; B: exponent of x+1 = 3 | |
Case 2
A: exponent of x−1 = 2; A: exponent of x+1 = 1; B: exponent of x−1 = 2; B: exponent of x+1 = 1 | |
Case 3
A: exponent of x−1 = 0; A: exponent of x+1 = 0; B: exponent of x−1 = 2; B: exponent of x+1 = 1 | |
Revise: My claim needs the condition… because…
Exit: Find the monic LCM of x−1 and (x−1)².
12 Polynomial factors and multiples · g10-12-03 · 5–10 min
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.
Case numbers match the projection buttons. Record the displayed conditions and your observation.
| Compare | What changed? What stayed? |
|---|
Case 1
Shared root r = 1; Constant u in x²+u = 1; Constant v in x+v = 2 | |
Case 2
Shared root r = 1; Constant u in x²+u = -1; Constant v in x+v = 1 | |
Case 3
Shared root r = 0; Constant u in x²+u = 0; Constant v in x+v = 0 | |
Revise: My claim needs the condition… because…
Exit: Why do A and B have the same GCD as B and A−B?
