Which polynomial has a multiple root at $x = 1$? (A) $x^5 - x^4 - x^2 + 1$ (B) $x^5 - x^4 - x - 1$ (C) $x^5 - x^3 - x^2 + 1$ (D) $x^5 - x^3 - x + 1$ - HSC - SSCE Mathematics Extension 2 - Question 2 - 2016 - Paper 1

1. Option A:
   Plugging $x = 1$ into $x^5 - x^4 - x^2 + 1$:

   $1^5 - 1^4 - 1^2 + 1 = 1 - 1 - 1 + 1 = 0$

   Now evaluate the derivative:
   $f'(x) = 5x^4 - 4x^3 - 2x$

   $$$$

eq 0$$
This option does not have a multiple root.

2. Option B:
   Plugging $x = 1$ into $x^5 - x^4 - x - 1$: $$$$

eq 0$$
This option does not have a root at $x = 1$.

3. Option C:
   Plugging $x = 1$ into $x^5 - x^3 - x^2 + 1$:
   $1^5 - 1^3 - 1^2 + 1 = 1 - 1 - 1 + 1 = 0$
   Now evaluate the derivative:
   $f'(x) = 5x^4 - 3x^2 - 2x$
   $f'(1) = 5(1^4) - 3(1^2) - 2(1) = 5 - 3 - 2 = 0$
   This option has a multiple root.

4. Option D:
   Plugging $x = 1$ into $x^5 - x^3 - x + 1$:
   $1^5 - 1^3 - 1 + 1 = 1 - 1 - 1 + 1 = 0$
   Now evaluate the derivative:
   $f'(x) = 5x^4 - 3x^2 - 1$

   $$$$

eq 0$$
This option does not have a multiple root.

Based on our evaluations, the polynomial in Option C: $x^5 - x^3 - x^2 + 1$ has a multiple root at $x = 1$, as both the polynomial and its derivative equal zero at this point.