Let P(x) be a degree 4 polynomial with P(1) = 1, P(2) = 5, P(3) = 9, P(4) = 13 and P(5) = 1. Determine the sum of the roots of P(x).

Step-by-step solution

1. 1

   Look for a pattern in the given values before doing any algebra. The first four outputs $1,5,9,13$ increase by $4$ each time, so they lie on the line

   $L(x)=4x-3$

   since $4(1)-3=1$, $4(2)-3=5$, $4(3)-3=9$, $4(4)-3=13$. Only the fifth value, $P(5)=1$, breaks the pattern (the line would predict $17$).

2. 2

   Subtract the pattern to manufacture known roots. Define

   $Q(x)=P(x)-(4x-3)$

   Then $Q(1)=Q(2)=Q(3)=Q(4)=0$. Subtracting a degree-$1$ polynomial cannot change the degree-$4$ leading term, so $Q$ is still a quartic — and a quartic with four known roots is completely determined up to a constant:

   $Q(x)=a(x-1)(x-2)(x-3)(x-4),\qquad a\neq 0$

3. 3

   Write $P$ explicitly. Undoing the subtraction,

   $P(x)=4x-3+a(x-1)(x-2)(x-3)(x-4)$

   This single formula already encodes four of the five conditions; the fifth one will pin down $a$.

4. 4

   Use $P(5)=1$ to find $a$. Substituting $x=5$:

   $1=4(5)-3+a(4)(3)(2)(1)=17+24a$

   $24a=-16\qquad\Longrightarrow\qquad a=-\frac{2}{3}$

5. 5

   Apply Vieta's formula, and watch $a$ cancel. For $P(x)=Ax^4+Bx^3+\dots$ the sum of the roots is $-\dfrac{B}{A}$. Expanding only the top two terms,

   $(x-1)(x-2)(x-3)(x-4)=x^4-(1+2+3+4)x^3+\cdots=x^4-10x^3+\cdots$

   so $A=a$ and $B=-10a$ (the trailing $4x-3$ touches neither coefficient). Therefore

   $-\frac{B}{A}=-\frac{-10a}{a}=10$

   The value of $a$ divides out. Computing $a=-\tfrac{2}{3}$ was a useful check, but the answer only needed $a\neq 0$, which is guaranteed because $P$ has degree $4$.

6. 6

   State the answer. The four roots of $P$ add up to

   $\boxed{10}$

   Sanity check with the explicit polynomial $P(x)=-\tfrac{2}{3}x^4+\tfrac{20}{3}x^3-\tfrac{70}{3}x^2+\tfrac{112}{3}x-19$: it reproduces all five given values, and $-\tfrac{20/3}{-2/3}=10$.