A polynomial p(x) satisfies the following:
- p(1) = p(3) = p(5) = -1
- p(2) = p(4) = 1
The minimum degree of such a polynomial is
A. 1
B. 2
C. 3
D. 4
GATE 2000 · Discrete Mathematics · Polynomials · medium
Answer: D. 4
Factor using the three roots of p(x)+1: Since p(1)=p(3)=p(5)=-1, the polynomial p(x)+1 vanishes at x=1,3,5. By the factor theorem: p(x)+1 = (x-1)(x-3)(x-5)*q(x), so p(x) = (x-1)(x-3)(x-5)*q(x) - 1 for some polynomial q(x).
Show degree 3 (constant q) is insufficient: Let q(x) = c (constant). Then p(x) = c(x-1)(x-3)(x-5) - 1. Condition p(2)=1: c(1)(-1)(-3)-1=1 => 3c=2 => c=2/3. Condition p(4)=1: c(3)(1)(-1)-1=1 => -3c=2 => c=-2/3. These are contradictory, so no degree-3 polynomial works.
Show degree 4 (linear q) works: Let q(x) = ax + b. From p(2)=1: (2a+b)(1)(-1)(-3)-1=1 => 3(2a+b)=2 => 6a+3b=2. From p(4)=1: (4a+b)(3)(1)(-1)-1=1 => -3(4a+b)=2 => 12a+3b=-2. Subtracting: -6a=4 => a=-2/3. Then 3b=2-6(-2/3)=2+4=6 => b=2. So q(x) = -2x/3 + 2, and p(x) = (-2x/3+2)(x-1)(x-3)(x-5)-1 has degree 4 and satisfies all five conditions.