Question 21

We evaluate $P(X) = X^5 + 4X^3 + 6X + 5$ using only the input $X$ and one temporary variable $T$.

1. Factor the polynomial to maximize reuse of $X$:

   $$P(X) = X(X^4 + 4X^2 + 6) + 5 = X((X^2 + 4)X^2 + 6) + 5.$$

2. Trace operations sequentially with single temporary variable $T$:

   • Operation 1 (multiplication): $T = X \times X$ ($T = X^2$)
   • Operation 2 (addition): $T = T + 4$ ($T = X^2 + 4$)
   • Operation 3 (multiplication): $T = T \times X$ ($T = X^3 + 4X$)
   • Operation 4 (multiplication): $T = T \times X$ ($T = X^4 + 4X^2$)
   • Operation 5 (addition): $T = T + 6$ ($T = X^4 + 4X^2 + 6$)
   • Operation 6 (multiplication): $T = T \times X$ ($T = X^5 + 4X^3 + 6X$)
   • Operation 7 (addition): $T = T + 5$ ($T = X^5 + 4X^3 + 6X + 5$)

3. Total count:

   • 4 multiplications (steps 1, 3, 4, 6)
   • 3 additions (steps 2, 5, 7)
     Total = $4 + 3 = 7$ arithmetic operations.

Hence, the minimum number of arithmetic operations required is $7$.