A polynomial p(t) is such that p(0) = -5, p(1) = 3 and p(-2) = -29. The minimum degree it should have is A. 1 B. 2 C. 3 D. 4

GATE 1997 · Discrete Mathematics · Polynomials · medium

Answer: B. 2

1. Test degree 1: p(t) = at + b: p(0) = b = -5. p(1) = a + b = 3 => a = 8. Check p(-2) = -16 - 5 = -21. But required value is -29. Since -21 ≠ -29, no degree-1 polynomial satisfies all three conditions.
2. Test degree 2: p(t) = at^2 + bt + c: p(0) = c = -5. p(1) = a + b - 5 = 3 => a + b = 8. p(-2) = 4a - 2b - 5 = -29 => 4a - 2b = -24 => 2a - b = -12. Adding (a+b=8) and (2a-b=-12): 3a = -4, so a = -4/3. Then b = 8 - (-4/3) = 28/3. Verify: p(-2) = (4)(-4/3) - (2)(28/3) - 5 = -16/3 - 56/3 - 5 = -72/3 - 5 = -24 - 5 = -29. ✓