Consider the set F of all polynomials whose coefficients are in the set of {0, 1}. Let q(x) = x3 + + 1. The number of polynomials p(x) in F of degree 14 such that the product p(x)q(x) is also in F is:
Hint 1: Start by analyzing the initial conditions and setting up the basic equations. p(x) q(x) = (x14 + …) (x3 + + 1).
Hint 2: Look for algebraic properties, symmetry, or geometric theorems to simplify. p(x) = x14 → 1 case.
Hint 3: Proceed with the final algebraic steps to solve the system. p(x) = x14 + x2.
Step 1: p(x) q(x) = (x14 + …) (x3 + + 1)
Step 2: p(x) = x14 → 1 case
Step 3: p(x) = x14 + x2
Step 4: => = 10, 9, 8, …, 10 → 11 case
Step 5: p(x) = x14 + x + x
Step 6: = 10, = 6,5,4,3,2,1,0
Step 7: = 9, = 5,4,3,2,1,0
Step 8: = 8, = 4,3,2,1,0
Step 9: = 7, = 3,2,1,0 25 cases
Step 10: = 6, = 2,1,0
Step 11: = 5, = 1,0
Step 12: = 4, =0
Step 13: p(x) = x14 + x + x + xr
Step 14: = 10, = 6, = 2,1,0
Step 15: = 10, = 5, = 1,0 6 cases
Step 16: = 10, = 4, = 0
Step 17: = 9, = 5, = 1,0
Step 18: = 4, = 0
Step 19: = 8, = 4, = 01 cases
Step 20: Hence, total case = 1 + 11 + 28 + 6 + 3 + 1
Step 21: = 50 cases.
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