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Lemma
Polynomials Difficulty 4

TMUA practice: Polynomials, problem 2

A TMUA-calibre polynomials problem at difficulty 4 of 5, with the full worked solution.

The question
Let P(X)=X4+a3X3+a2X2+a1X+a0P(X) = X^{4} + a_{3} X^{3} + a_{2} X^{2} + a_{1} X + a_{0} be a (monic) polynomial with real coefficients satisfying P(0)=1, P(1)=2, P(2)=3, P(3)=4P(0) = 1,\, P(1) = 2,\, P(2) = 3,\, P(3) = 4. What is the value of P(4)P(4)?
A 55
B 2424
C 2929
D not determinable from the given data\text{not determinable from the given data}

The correct answer is highlighted. C

Worked solution

Define   Q(X)=P(X)−(X+1)\;Q(X) = P(X) - (X + 1). The conditions become Q(0)=Q(1)=Q(2)=Q(3)=0Q(0) = Q(1) = Q(2) = Q(3) = 0, so X=0,1,2,3X = 0, 1, 2, 3 are four roots of QQ.

QQ has degree 44 (since PP is monic degree-44 and X+1X + 1 is degree-11). With four roots accounted for, QQ factors as

Q(X)  =  X(X−1)(X−2)(X−3).Q(X) \;=\; X(X - 1)(X - 2)(X - 3).

So   P(X)=(X+1)+X(X−1)(X−2)(X−3)\;P(X) = (X + 1) + X(X - 1)(X - 2)(X - 3).

Evaluate at X=4X = 4:

P(4)  =  (4+1)+4⋅3⋅2⋅1  =  5+24  =  29.P(4) \;=\; (4 + 1) + 4 \cdot 3 \cdot 2 \cdot 1 \;=\; 5 + 24 \;=\; 29.

This is option C.

Why the other options fail.

  • A, 5=4+15 = 4 + 1: follows from extrapolating P(n)=n+1P(n) = n + 1 linearly, but the quartic correction kicks in at X=4X = 4.
  • B, 2424: the value of the correction 4⋅3⋅2⋅14 \cdot 3 \cdot 2 \cdot 1 alone, without the linear part.
  • D: the data are sufficient because monic-quartic with four prescribed values is unique.

The lesson: when a degree-nn polynomial is given at nn consecutive integers and its leading coefficient is specified (here = 11), it is uniquely determined. Use the ‘subtract the obvious interpolant’ trick.